5. Dynamics of rigid bodies

Introduction

Motorcycle Racing

In order to make a curve with a bicycle or a motorcycle there must be sufficient friction between the tires and the road, because the frictional force provides the centripetal force that makes the bike follow the curved trajectory. But the frictional force also makes the rider and the bike rotate sideways in the direction opposite to the center of the curve; to counteract that rotation the rider tilts with the bike, making their weights produce a tendency to rotate in the opposite direction. The tendency to rotate produced by a force is called moment and will be studied in this chapter. The high speeds in motorcycle races imply larger tilt angles. To make the motrocycle tilt more, the rider first turns the wheel in the opposite sense of the curve and moves his body away from the motorcycle in the direction of the center of the curve.

5.1. Sliding vectors

The vectors introduced in chapter 2 are free vectors, which are considered equal if they have the same magnitude, direction and sense, regardless of the point of the space where they are. In the case of forces, it is not enough to know their magnitude, direction and sense. For example, when a force is applied to a door to close it, besides the magnitude, direction and sense of the force, the point at which that force is applied will also be important. The further away from the hinges the force is applied, the easier it is to close the door. The force required to close the door would be very high if it was applied at a point near the hinges.

Suppose we want to move a chair to another place, lifting it with just one hand. If the chair weights 40 N, and its center of gravity is at the point C, as shown in figure 5.1, we would probably put our hand at a point P on the back of the chair and apply an upward force of 40 N to lift it. As the chair moves up, it will rotate until it reaches an equilibrium position as shown on the right-hand side of the figure, where the weight and the force applied by the hand are in the same vertical line. If we could apply the upward force in a point in the same vertical line as the center of gravity C, the chair would go straight up without rotating.

Lifting a chair

Figure 5.1: A chair of weight 40 N being lifted with an upward force.

We conclude that to predict the effect of a force it is necessary to know its magnitude, direction, sense and also its line of action, which is the straight line passing through the point where the force is applied, in the direction of the force. A force produces different effects if it is moved to a different line of action, even if its magnitude, direction and sense are kept equal.

Sliding Vectors

Figure 5.2: Three equivalent points.

These kind of vectors, with a specific line of action, are called sliding vectors. The point where they are applied is not important, as long as it is in their line of action. In the example of the chair being lifted, the same upward force could have been applied at points Q or R in figure 5.2, which are in the same vertical line as P, and the effect would have been the same. The chair would rotate in the same direction, with the same angular acceleration. Therefore, it is equivalent to assume that is applied in P, Q, R or any other point in its line of action. However, even if the initial effect is the same, once the chair starts to rotate, the higher the point where is applied, the more the chair will rotate until the line of action of its weight coincides with the line of action of , reaching the equilibrium position.

Whenever it was necessary to add forces in chapter 4, we assumed that they could be moved freely and added together as free vectors, using the parallelogram rule. In the following sections it is shown that this sum of forces as if they were free vectors is not wrong, as long as the rotation effect introduced when moving a force to a different line of action is also taken into account. The translation of a body is determined by the resultant force, which can be obtained adding all external forces as free vectors. The rotation, or lack of it, is also determined by the external forces but regarded as sliding vectors and not as free vectors..

5.2. Superposition of forces

Sum of collinear forces

Figure 5.3: Collinear forces.

To determine how forces and other sliding vectors must be added, let us start by the simplest case, when the forces to be added have the same line of action. For example, if we are going to lift a book that is resting on a horizontal table, as in figure 5.3, in order to apply an upward force in the same vertical line as the center of gravity we could drill a hole through the book, going through the center of gravity, and pass a string through the hole attaching it to a washer on the bottom of the book.

When we pull up the string with a force , that force will be collinear with the weight , i.e. their lines of action are the same, so the two forces can be moved to a common point in their line of action and added as free vectors. If is bigger than , the book accelerates upwards without rotating; is then reduced to to maintain a constant speed and it is then reduced even more to make the book stop at some height.

If instead of drilling a hole through the book we pass a strip of paper under it, as shown in the left side of figure 5.4, and pull up the two ends of the strip, keeping the book horizontal, we would be applying two forces and as shown in the middle of the figure.

Addition of concurrent forces
Figure 5.4: Addition of concurrent forces.

Those two forces are called concurrent, because their lines of action cross each other at a point P. We can then move each force along its line of action, to the point P, and add them there as free vectors to give the resultant force , as shown in the right side of figure 5.4. Notice that, if the book is lifted keeping it horizontal, it does not rotate so the resultant force is collinear with the weight and it is added to it as in the case shown in figure 5.3.

When the lines of action of two forces and are parallel, as is the case in figure  5.5, we can use the following procedure to add them: two additional collinear forces and are added, in a direction perpendicular to the lines of action of and , as shown in the middle of the figure. That does not alter the system, since the sum of those two additional forces is zero. The force is then added together with , in their common point, giving a force , and is added together with , in their common point, giving a force . These two new forces and are now concurrent and can be added in their common point P (right side of figure  5.5) giving the resultant force .

Sum of parallel forces
Figure 5.5: Addition of parallel forces.

Notice that the resultant force is also in the same direction as the original forces and with magnitude equal to the sum of the magnitudes of and , since . But the line of action of the resultant force will be another line, parallel to the lines of action of and and in between them.

Line of action of the resultant force

Figure 5.6: Line of action of the resultant.

To determine the distances and between the lines of action of the parallel forces and , and their resultant , we see in figure 5.6 that the segment ) is given by,

(5.1)

On the other hand, the angles and are also the angles between and , and between and ; therefore,

(5.1)

and substituting the tangents of the angles into equation 5.1 we get a result called law of the lever:

(5.3)

To lift the book with two vertical forces, without making it rotate, it would be necessary to apply two forces with magnitudes inversely proportional to their distances to the center of gravity, as shown by equation 5.3.

5.3. Moments and couples

The law of the lever can be explained by introducing the concept of moment of a force about a point. The moment of a force about a point O is defined as the product of the magnitude of the force by the distance from point O to the line of action of :

(5.4)

is called the moment arm of the force and the moment about a point is also called torque. The moment measures the rotational effect produced by the force when point O is fixed. The further away the action line of the force is from point O, the bigger the rotation effect it produces. This explains why it is more difficult to close or open a door if the force is applied too close to the hinges: the moment of the force about the hinges is small when its line of action is close to the rotation axis of the door.

Moment of a force

Figure 5.7: Moment of a force.

If is the position vector of the point P where the force is applied, relative to point O, the moment arm of the force about point O is equal to , where is the angle between the vectors and (figure  5.7). We then conclude that the moment of the force about point O is equal to,

(5.5)

It should be noted that ( ) is the component of the force in the direction perpendicular to the position vector , that is, the moment of the force about O is also equal to the product of the distance , from its point of application to point O, times the component of the force perpendicular to . The moment produced by a force is due solely to its component perpendicular to the segment from the reference point to the point where the force is applied.

Equation  5.5 shows that the moment of a force is also equal to the magnitude of the cross product between the position vector and the force. It is convenient to define the vector moment:

(5.6)

The vector produces a rotation about O on a plane perpendicular to and in the sense given by the right-had rule. In figure 5.7 the moment is a vector that points outside of the figure and can be represented by an arc with an arrow in the sense of the rotation.

A couple

Figure 5.8: Couple.

Consider two equal and opposite forces and , acting on different lines of action, as shown in figure 5.8. Their resultant cannot be obtained by the method described in the previous section, but we can see that is zero and they produce no translation in any direction, only rotation. The total moment about a point O is the sum of the moments of the two forces:

(5.7)

The two position vectors of the points Q and P depend on the choice of the origin O, but their difference does not change if the origin O is moved to another point.

This means that the two opposite forces can be applied in any part of the body, in any direction and with any magnitude, as long as they are equal and opposite and the product of their magnitude by the distance between their lines of action is the same. The total moment they produce is called a couple and it is represented by , without any index:

(5.8)

The couple of the two forces in figure 5.8 is a vector pointing out of the figure, represented by the circular arc with an arrow in the counterclockwise direction.

A force applied at a point P can be moved to another point Q, outside its line of action, using the procedure illustrated in figure 5.9: we add two equal and opposite forces and in points P and Q and we also add a couple , with the same magnitude of the couple produced by and , but in the opposite direction; then cancels and so the system remains the same. In the case of figure 5.9, Must be clockwise and with a magnitude equal to times the distance from Q to the line of action of the original force; or, as a vector, . We are then left with two equal and opposite forces at point P that cancel out, the force at point Q and the couple which is equal to the moment of the original force, in P, about point Q.

Moving a force to another line of action
Figure 5.9: Procedure to move a force from point P to another point Q.

To add all the external forces and couples acting on a body, we can move all the forces to a point Q, including the corresponding couple that accounts for having moved each force to a different line of action. We then add all the couples and forces in Q giving the resultant couple and the resultant force .

When the lines of action of all the forces are in the same plane, it will be convenient to define two of the coordinate axes in that plane, for example and , and the origin at the point where the forces will be added. Thus, the moment of each force about the origin adds a couple to the resultant couple, with only component, given by the determinant:

(5.9)

where and are the coordinates of the point where the force is applied. If there are external forces and couples (some of them due to the displacement of the forces to O and some could be external couples) the resultant force and couple are then

(5.10)

The resultant force can then be moved to another line of action, a distance from its line of action in Q, in order to eliminate the resultant couple and leaving just the resultant force.

5.4. Rigid bodies in equilibrium

If the resultant couple and force are both zero in a point, they will also be zero in any other point and the rigid body is said to be in equilibrium. The equilibrium state can be static, when the body is at rest, or kinetic, when the body has uniform linear motion without rotation. Thus, the conditions for a rigid body to be in equilibrium are that the sum of the external forces is zero and the sum of the moments of the forces, relative to any point, is also zero.

Example 5.1

The car in the figure is stopped on a horizontal road. The car weighs 9000 N and its center of gravity, C, is 60 cm behind the front wheels axle and 120 cm in front of the rear wheels axle. Determine the normal force on each tire.

A car on a horizontal road

Forces in a car at rest

Resolution. The car is in equilibrium, since it is at rest. The figure on the right shows the free-body diagram for the car; is the sum of the two normal forces on the front tires and is the sum of the normal forces on the rear tires. If there are frictional forces in the tires, they can not be calculated when we consider the whole car as a single rigid body; all we can conclude is that the sum of the frictional forces in the four tires is zero. The condition for the sum of the vertical forces to be zero is:

To find the value of these two variables, we must also use the condition that the sum of the moments of all forces, about any chosen point, must be zero. The sum of the moments about point A is:

This value could then be substituted in the condition for the sum of the vertical forces, but it is also possible to use the condition that the sum of moments about point B is also zero:

Assuming that the center of gravity is at the same distance from the left and right tires, the normal forces on the two front tires would be the same and therefore the normal force on each front tire would be  N. And in each rear tire the normal force would be  N.


If the car were moving with constant velocity, it would still be in equilibrium but we would then have to take into account the air resistance which is a force pointing to the right and a little downwards. There would be frictional forces in the tires, the sum of which would be equal and opposite to the horizontal component of the resistance force of the air. Since the frictional forces and the air resistance force act on different lines of action, they would produce a couple that would make the car rotate, increasing the normal forces in the rear tires and diminishing the normal forces in the front tires. To compute that couple, it would be necessary to know the drag coefficient , the speed of the car and the point of application of the resultant resistance force.

5.5. Center of mass

The mass in a rigid body is continuously distributed in a volume. If the total mass of the body is and is the infinitesimal mass that exists in the neighborhood of a point in the body, then

(5.11)

Where the integral is a triple integral, within the volume occupied by the solid, since is the product of the density times the infinitesimal volume .

The position vector of the center of mass, , is defined as the mean value of the positions of the parts of the body, weighed by their mass; namely,

(5.12)

Example 5.2

Find the position of the center of mass of the homogeneous solid shown in the figure.

Triangular wedge

Resolution . The volume of the solid is delimited by the 5 planes , , , and .

The infinitesimal area Is equal to the density times the infinitesimal volume, which in cartesian coordinates is . We begin by calculating the total mass from equation 5.11:

Since the body is homogeneous, is constant. In Maxima, the three integrals must be calculated sequentially; the variable p will represent the density

(%i1) integrate (p, z, 0, c*(1 - x/b))$
(%i2) integrate (%, x, 0, b)$
(%i3) m: integrate (%, y, 0, a);
(%o3)     

Although the intermediate results have not been shown, they are stored in the %o1 and %o2 variables.

To compute , the same volume integral is repeated, changing the integrand from to ( )

(%i4) r: [x, y, z]$
(%i5) integrate (p*r, z, 0, c*(1 - x/b))$
(%i6) integrate (%, x, 0, b)$
(%i7) rcm: integrate (%,y,0,a)/m;
(%o7)     

We conclude that the position vector of the center of mass is: .


In every rigid body there is always a single point which is the center of mass. If the origin is chosen exactly at the center of mass, the integral in equation 5.12 will be zero for each of the three components:

(5.13)

The integrals in 5.13 are all zero only if the origin is at the center of mass.

Differentiating the two sides of equation 5.12, with respect to time, gives the expression of the velocity of the center of mass:

(5.14)

That is, the velocity of the center of mass is the average of the velocities of all parts of the body, weighted by their masses.

And differentiating equation 5.14 with respect to time we obtain the acceleration of the center of mass:

(5.15)

Which is the mass-weighted average of the accelerations of all the parts of the body.

If the reference in which the accelerations are measured is an inertial reference frame, the product is equal to the total force acting on the part with mass :

(5.16)

Note that whenever a rigid body is accelerating, there must be an infinitesimal force applied to each part of the body, making them all move together. In most points that force is due solely to the internal forces of contact between the parts of the body, forces that are triggered throughout the body by the action of external forces , , , applied in points: 1, 2, , , where the force includes the contact forces plus a external force. The differential gives the variation of the total force throughout the body.

Substituting the expression  5.16 into equation  5.15, we conclude that,

(5.17)

In the integral on the left-hand side, for each internal contact force that exists at one part, there will be another equal but opposite force at another neighboring part, due to the law of action and reaction. Therefore, all the internal forces of contact will be eliminated and the integral will give the sum of the external forces, , , , , Which is the resultant force on the rigid body. As such, equation 5.17 leads to,

(5.18)

This important result is the law of motion for the translation of a rigid body:

The motion of the center of mass of any rigid body with mass is equal to the motion that would have a point particle with mass under the action of a force equal to the resultant force on the rigid body.

Remember that the resultant force is obtained by adding all external forces as if they were free vectors. If the resultant force is zero, the center of mass will either be at rest or in a state of uniform rectilinear motion, but other points on the rigid body may have more complicated motions.

In the case of the weight, the external force is applied in all parts of the rigid body and the resultant force is . Hence, equation 5.17 gives,

(5.19)

If the acceleration of gravity is the same in all points of the body, the integral on the left-hand side will be equal to and we conclude that the acceleration of the center of mass equals the acceleration of gravity and that the center of gravity, i.e. the point where the total weight acts, coincides with the center of mass. For rigid bodies with size much smaller than Earth, the acceleration of gravity can be considered constant throughout the whole body and the center of mass is at the same point as the center of gravity.

Consider, for example, a triangular blade. Hanging it by one of the vertices, it will begin to oscillate until it stops in a position where the center of gravity is in the same segment of vertical line that passes through the vertex; by tracing this segment in the triangle and repeating the procedure for the other two vertices, the point where the three segments intersect will be the center of gravity and center of mass. If the density of the triangle is equal at all points, each of the vertical segments will be the median that divides the triangle into two parts with the same area and mass; therefore, the center of mass is at the intersection of the three medians. In bodies with symmetrical shapes and constant density, the center of mass lies in the geometric center. 5.10 shows three other examples.

Center of mass centers of some symmetrical bodies
Figure 5.10: Center of mass of 3 objects with constant density: a sphere, a cylinder and a parallelepiped.

5.6. Translation without rotation

In a rigid body moving along any path without rotating, the acceleration of all points is the same at every instant, equal to the acceleration of the center of mass, which is equal to the resultant force divided by the mass of the body. As the body does not rotate, the resultant couple about the center of mass should be zero. It should be noted that resultant couple is zero only about center of mass and not about other points. The resultant couple about an arbitrary point P, equals the couple of the resultant force, applied at the center of mass.

Example 5.3

The same car of example 5.1, accelerates during 20 s, starting from rest and reaching a final speed of 60 km/h. Knowing that the center of gravity is 57 cm high above the ground, and assuming that the acceleration is constant, determine the normal forces and the frictional forces in the tires, at the instant when it starts moving.

Resolution. When the car starts moving there is no air resistance force, because its speed is zero. Therefore, the only horizontal forces, responsible for the acceleration, are the static frictional forces between the tires and the road. The following figure shows the free-body diagram of the car.

Forces in a car accelerating

is the sum of the two normal forces in the two front tires, is the sum of the normal forces in the rear tires and is the sum of the frictional forces in the four tires. Since the tangential acceleration is constant, it is then equal to increase of the velocity, divided by the time it took to increase:

The law of motion for translation leads to the following equations:

Since the car is not rotating, the resultant couple about the center of mass must be equal to zero. The weight does not produce any moment about the the center of mass. The moments of and about the center of mass are in the clockwise direction and the moment of about the center of mass is in the counterclockwise direction. The resultant couple about the center of mass is then:

The solution of the system of the 3 equations leads to,

The total frictional forces in the four wheels is 765 N, the normal force on each front tire is 2879 N and the normal force on each back tire is 1621 N.

5.7. General motion of the rigid body

The dynamics of the rigid body consists of the study of the effects of external forces and couples on the variation of its six degrees of freedom. The trajectory of any point in the body, used as reference point, gives the variation of three of these degrees of freedom. The remaining 3 degrees of freedom are 3 angles. In the spinning top shown in figure 5.11, the two angles and define the direction of the axis of the top; a third angle, , is related to the rotation of the top with respect to its axis. In this case two of the angles, and , vary with time and therefore, there are two angular velocities and .

Spinning top
Figure 5.11: The 3 degrees of freedom in the rotation of a rigid body.

Since the axis of the spinning top is not in a vertical position, its weight and the normal force on the tip of the top produce a couple in the direction in which increases, but as the top is rotating in the direction shown for , the axis of the top does not fall but moves in the circle indicated for the angle .

The general motion of a rigid body is easier to study using lagrangian dynamics, instead of the vector dynamics used in this chapter. We will conclude the chapter considering a simpler case in which the axis of rotation does not move, so the is only one angle changing with time.

5.7.1. Rotation with fixed axis

When the axis of rotation of a rigid body remains fixed relative to an inertial frame, Newton's second law will be valid for the accelerations measured in the frame of the rigid body. Thus, equation 3.24 allows us to calculate the force acting on each part with differential mass

(5.20)

Each of these forces produces a moment about the origin, but since the rigid body can only rotate around the fixed axis , it is only relevant to compute the component of the moment, obtained using only the radial component of the position vector:

(5.21)

By integrating in the volume of the rigid body the component of the resulting moment is then,

(5.22)

The angular acceleration was placed outside the integral, because it is the same at all points of the rigid body. The integral on the right-hand side,

(5.23)

Is called moment of inertia of the rigid body, relative to the axis.

In the integral , all moments of the internal forces of contact will cancel, as a consequence of the law of action and reaction, and the result will be resultant couple of the external forces, , , , . Thus, equation  5.22 leads to the law of motion for rigid body rotation around a fixed axis:

(5.24)

The angular acceleration of a rigid body around a fixed axis, times its moment of inertia around that axis, is equal to the component of the resultant couple along that axis.

Example 5.4

Determine the moment of inertia of a homogeneous cylinder of radius and height , around its axis.

Resolution. It is convenient to use cylindrical coordinates with the axis at the axis of the cylinder. The volume of the cylinder is defined by the conditions: , and (we use for the cylindrical coordinate, to avoid confusion with the cylinder radius).

The differential element of volume in cylindrical coordinates is ( ) and hence, , where Is the density. The moment of inertia is,

Note that the mass of the cylinder is obtained by the integral:

Thus, the expression for the moment of inertia is:


The role played by the moment of inertia in rotations is similar to the role of the mass in translation motions. The bigger the moment of inertia, the less the angular acceleration that the resultant couple will produce. Observe the similarity of equation  5.24 with Newton's second law.

Table  5.1 shows the expressions of the moment of inertia of some solids around the axes that pass through their center of mass. The moment of inertia is always directly proportional to the mass of the body. The radius of gyration is a typical length for each geometrical shape, defined as the square root of the moment of inertia divided by the mass:

(5.25)
Table 5.1: Moments of inertia of some solids with constant density around axes passing through the center of mass.
BallCylinder

Parallelepiped

Ball
Cylinder
Parallelepiped
Axis 1: 
Axis 2: 

The moment of inertia around an axis through the center of mass can be used to obtain the moment of inertia around any other parallel axis out of the center of mass, using the parallel-axis theorem :

(5.26)

where is the mass of the body and is the distance between the two parallel axes. It is also possible to compute the moment of inertia of a body by adding the moments of inertia of the various parts of it, since the integral 5.23 can be written as the sum of the integrals in the various parts. The moment of a sufficiently thin bar of length can be obtained from table 5.1, either for the expression for a cylinder, around axis 1 and in the limit , or from the expression for a parallelepiped with and

A pulley with a shaft attached to a fixed base is an example of a rigid body with a fixed rotation axis. If the pulley is homogeneous, the center of mass will also be on the axis of the shaft. Figure 5.12 shows a pulley of mass and radius , with a string that moves with the pulley without sliding on its surface. The external forces and couples are the weight, , the tensions in the string in both sides of the pulley, and , the contact force on the shaft, and a couple produced by friction between the shaft and the pulley, in the direction opposite to the rotation of the pulley (the pulley was assumed rotating clockwise in figure 5.12).

Forces in a fixed pulley pulley
Figure 5.12: External forces and couples on a pulley.

The weight of the pulley and the contact force do not produce moment around the axis. Since the pulley is a cylinder, using the expression for the moment of inertia in table 5.1, the equation for the resultant couple is (assuming the angular acceleration is in the clockwise sense),

(5.27)

When friction on the shaft can be ignored,

(5.28)

where is the tangential acceleration of a point on the string. Note that, regardless of the radius of the pulley, when its mass is much smaller than and , it may be assumed that the tension is equal on both sides of the rope.

Questions

(To check your answer, click on it.)

  1. The cartesian components of a force are . In which of the positions on the list should the force be applied to produce a clockwise moment about the origin?
  2. Two external forces are exerted on a wheel, as shown in the figure. Compute the resultant moment about point O, in units of N·m.
    External forces on a wheel
    1. 0.57
    2. 1.05
    3. 4.35
    4. 5.67
    5. 6.15
  3. A metal part with constant density and mass is constructed with two cylinders of the same height, but two different radii , glued one over the other so that their axes are aligned. Find the expression of the moment of inertia of the part around its axis of symmetry.
  4. Two children with masses 30 kg and 45 kg are sitting on both sides of a seesaw. If the heavier child is sitting 1.2 m from the pivot point of the seesaw, how far from that point should the other child sit to keep the seesaw balanced?
    1. 1.5 m
    2. 0.8 m
    3. 1.8 m
    4. 1.2 m
    5. 0.98 m
  5. If an object is divided into two parts, making a vertical cut along a line that passes through its center of gravity, which of the statements about the two pieces obtained is true?
    1. They must have the same mass.
    2. They may have different masses.
    3. They should have the same weight.
    4. They must have the same area.
    5. They must have the same volume.

Problems

  1. The hammer in the figure is placed over a block of wood of 40 mm of thickness, to facilitate the extraction of the nail. If a force of 200 N (perpendicular to the hammer) is required to extract the nail, find the force on the nail and the force at point A while the nail starts to be removed in the position shown. Assume that the weight of the hammer can be neglected, compared to the other forces and that there is enough friction in A to prevent the hammer from sliding.

    Hammer extracting a nail
  2. A front-wheel drive car accelerates evenly from rest reaching a speed of 100 km/h in 11 seconds. If the car's weight is 9750 N, find the normal reactions and the frictional force on each tire, in the moment when it starts moving. What is the minimum value that should have the coefficient of static friction between the tires and the road so that the car can achieve this acceleration?

    Center of gravity of a car
  3. A 45 kg cabinet, mounted on wheels that let it roll freely on the floor, is accelerated by an external force of 310 N.
    1. Find the maximum and minimum values ​​that the height can have for the cabinet to accelerate without the wheels losing contact with the ground.
    2. Find the cabinet acceleration when is between the minimum and maximum values ​​obtained in part a.

    Cabinet with wheels being pushed
  4. Using integration in the volume of the solid, prove the result of table 5.1 for the moment of inertia of a rectangular parallelepiped with axis of rotation perpendicular to one of its faces and passing through the center of mass.
  5. A homogeneous cylindrical wooden beam, with diameter of 48 cm, length of 3 m and mass of 100 kg, is hung horizontally by means of two 2 m ropes, as shown in the figure. The beam is released from rest in the position where each rope makes an angle of 60° with the horizontal. Determine the tension and angular acceleration of each rope at the exact moment the beam is released from rest.

    Hanging trunk
  6. The ladder in the figure rests on a horizontal surface (point A) and a vertical wall (point B). The coefficient of static friction between the ladder and the horizontal surface is , while the friction of the ladder with the vertical wall is negligible. Assuming that the center of gravity of the ladder is at half its length, determine the minimum value of , to ensure that the ladder remains at rest.

    Staircase leaning against a wall
  7. The mass of the trailer in the figure is 750 kg and is being hitched by a car linked to the trailer at point P. The road is horizontal and the two identical tires can be considered as one, with a single normal reaction and negligible frictional force; the air resistance will also be neglected. (a) Compute the normal force in the tires and the vertical force at point P, when the speed is constant. (b) When the car is accelerating, with  m/s2, the force at P will have horizontal and vertical components. Compute these components and the normal force in the tires (the moment of inertia of the wheels and the friction with the road are negligible).

    Trailer
  8. The homogeneous rectangular box in the figure is attached to two hinges that let it turn covering the window, or move to the horizontal position shown in the figure, to provide shade during the day. The chain that holds the box in the horizontal position suddenly breaks and the box falls hitting the wall. Determine the angular velocity with which the box hits the wall, neglecting the friction in the hinges and the air resistance.

    Window with sunshade
  9. The airplane in the figure, with a total mass of  kg, lands on a horizontal track. Point C represents the center of gravity. At the instant the speed is 210 km/h (to the right), the pilot turns the turbines in reverse mode, producing the constant force shown in the figure, and after 580 m on the track the speed decreases to 70 km/h. During that motion, the frictional forces in the tires and the air resistance can be neglected, compared to which is much higher. Compute the normal force on the front wheel.

    Airplane landing
  10. A 91 kg man pulls a truck on a horizontal road, with constant speed, using a rope tied to his back. The figure shows the relative positions of the center of gravity of the man, C, the point of contact of his foot with the ground, A, and of the point where the rope is attached to his back, B.
    1. Compute the magnitude of the tension on the rope.
    2. Draw a diagram of the forces you think might be acting on the truck.

    Man pulling a truck
  11. To test its brakes, a bicycle was placed with the wheels in the air and the front wheel was made to rotate freely, as shown in the figure. The time taken for the wheel to make 10 turns was measured giving 8.2 s (assume that during those 8.2 s the angular velocity remained constant). Immediately afterwards, the brakes were applied and the wheel took 2.9 s until it stopped completely. The figure shows the frictional force between the friction pads and the rim, which is tangent to the rim and applied at a distance of 27.1 cm from the wheel axis.
    1. Assuming that the force is constant, the angular acceleration it produces will also be constant; compute the angular acceleration.
    2. Determine how may turns the wheel rotated from the moment the brakes were applied until it stopped.
    3. Knowing that the moment of inertia of the wheel around its axis is 0.135 kg·m2, find the magnitude of the force .

    Bicycle wheel
  12. The 1.5 kg cylinder in the figure descends vertically, causing the 5 kg block to accelerate on the horizontal table. The pulley can be considered a uniform disk with mass of 0.4 kg. The string rotates with the pulley without sliding on its surface. The coefficient of kinetic friction between the block and the table is 0.2. Determine the value of the acceleration of the block and cylinder, neglecting the friction in the pulley's shaft, the mass of the string and the resistance of the air.

    System with block, pulley and cylinder
  13. A wheel A, with radius of 20 cm, mass of 600 g and a radius of gyration of 12 cm (relative to its axis), is connected to a motor that exerts a couple  N·m on it, in the direction shown in the figure. Wheel B, with radius of 15 cm, mass of 400 g and radius of gyration of 10 cm (relative to its axis) remains in contact with wheel A, without slipping. Wheel B is attached to its shaft by means of a bearing, which causes the friction in the shaft to be negligible. Determine the angular acceleration of wheel B. Hint: analyze the free-body diagrams of the two wheels, keeping in mind that there is frictional force between them.

    Wheels in contact
  14. To measure the moment of inertia of a pulley with 5 cm of radius, one end of a string was glued to it, then rolled around it and the other end was connected to a cylinder of mass  kg. The cylinder descends freely unrolling the string and making the pulley accelerate with angular acceleration  Rad/s2. Knowing that the friction in the pulley shaft produces a constant couple  N·m, in the direction indicated in the figure, find the moment of inertia of the pulley around its axis.

    Shaft pulley with shaft friction
  15. A cylindrical object has mass distributed uniformly between and (with ), with a cylindrical orifice of radius , as shown in the figure. Find the expression of the moment of inertia around the axis of the cylinder. Hint: Determine the mass of the complete cylinder, without orifice, in terms of , and the mass of the cylinder which is removed when the orifice is made; the moment of inertia is equal to the moment of inertia of the complete cylinder (see table 5.1 ), plus the moment of inertia of the cylinder that has been removed, assuming that the mass of the latter is negative!

    Hollow cylinder

Answers

Questions: 1. E. 2. D. 3. C. 4. C. 5. B.

Problems

  1. The nail exerts a downward force of 1000 N. (N)
  2. Front Tires:  N,  N. Rear tires:  N, (Assuming the rear wheels are perfectly free). The minimum coefficient of static friction is 0.416.
  3. (a) Minimum height 38.6 cm, maximum height 135.4 cm (b) (m/s2)
  4. In this case and the volume of the solid is defined by , , .
  5.  N,  N,  Rad/s2
  6. 0.21
  7. (a)  N,  N. ( b )  N,  N,  N.
  8. 5.274 s−1
  9.  N.
  10. (a) 559 N. (b) The forces acting in the truck are the tension of the rope, , the weight, , the normal forces in the tires, and and the frictional forces on the tires, and :

    Forces on a truck
  11. (a) 2.64 s2. (b) 1.77 turns. (c) 1.32 N.
  12. 0.7313 m/s2.
  13. 67.72 s−2
  14. 0.0062 kg·m2
Wrong

(Click to continue)

Wrong

(Click to continue)

Wrong

(Click to continue)

Wrong

(Click to continue)

Correct

(Click to continue)

Wrong

(Click to continue)

Wrong

(Click to continue)

Wrong

(Click to continue)

Correct

(Click to continue)

Wrong

(Click to continue)

Wrong

(Click to continue)

Wrong

(Click to continue)

Correct

(Click to continue)

Wrong

(Click to continue)

Wrong

(Click to continue)

Wrong

(Click to continue)

Wrong

(Click to continue)

Correct

(Click to continue)

Wrong

(Click to continue)

Wrong

(Click to continue)

Wrong

(Click to continue)

Correct

(Click to continue)

Wrong

(Click to continue)

Wrong

(Click to continue)

Wrong

(Click to continue)