Solved Problems

5. Dynamics of rigid bodies

Problem 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

Problem 5.1

In order to extract the nail without bending it, the hammer must be used in a way that it makes a vertical upward force over the nail. The reaction to that force will the force that the nail exerts on the hammer, with the same magnitude , but pointing downward, as shown in the free-body diagram on the side. The weight of the hammer has been ignored and the contact force in A was separated into its two cartesian components and .

If the nail is extracted without making the hammer accelerate, the tangential and normal acceleration of the hammer are both zero; therefore, the hammer is in equilibrium. Hence, the sum of the moments of the forces about any point is zero. Only the 200 N force and produce moment about A, so the some of moments about that point is:

The sum of the horizontal components of the forces must be also equal to zero

And so must be the sum of the vertical components

Problem 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 beam

Problem 5.4

The figure shows the free-body diagram for the beam, where and are the tensions in the two ropes and the center of mass is at the centroid of the beam. since the trunk remains in a horizontal position as it oscillates, it has no angular velocity; the motion of the beam is then a translation with no rotation, with all the points having the same circular trajectory, perpendicular to the two ropes.

It is convenient to use a system of tangential and normal axes as shown in the figure, with origin at the center of mass. The equations for this system are 3: the sum of the tangential components of the external forces is equal to the mass times the tangential acceleration, the sum of the normal components of the external forces is equal to the mass times the normal acceleration, which is zero at the instant when the beam is released from rest, and the the sum of the moments of the external forces about the origin (center of mass) is equal to zero.The first two equations are then:

And the sum of the moments about the origin is,

These three equations can be solved in Maxima:

(%i1) float (solve ([9.8*cos(%pi/3)=at, T1+T2-980*sin(%pi/3)=0,

At points A and B, the two ropes have the same acceleration as the beam, equal to 4.9 m/s2. Therefore, the angular acceleration of the two ropes is the same and equal to the acceleration of the beam divided by the length of the ropes (2 m), namely

Problem 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 being hitched

Problem 5.7 a

(a) The figure shows the free-body diagram for the trailer. since the velocity is constant, the trailer is in equilibrium. The normal force, , can be computed adding moments of the forces around point P and equating the sum to zero

And the force in P, , is found from the of moments about the point of contact of the tire with the road

Problem 5.7 b

(b) The figure on the side shows the free-body diagram of the trailer while it is accelerating. Since the acceleration is in the direction (horizontal) and the trailer doesn't rotate, the sum of the components of the forces must be equal to , the sum of the components of the forces must be zero and the sum of the moments around the center of mass must be zero too:

The solution of the second and the third equations is the following

(%i2) float (solve ([Rn+Fy-7350=0, 33*Rn-95*Fy-40*1500=0]));

Problem 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.

Plane slowing down

The following figure shows the free-body diagram of the plane, where is the normal force on the front wheel and is the sum of the normal forces on two rear wheels. The motion of the plane is a translation without rotation.

Problem 5.9

The conditions for the sums of the components of the forces and the moments around the center of mass are the following:

Since the force remains constant, the first equation implies that the acceleration is also constant; therefore, the differential equation that relates the acceleration with the velocity and the position can be integrated

The minus sign indicates that the acceleration is opposite to the velocity. Hence,

Substituting that value into the equation of the sum of moments, the system of two equations for and could be solved. However, an easier way to find consists on writing the equation for the sum of the moments about point B:

Notice that the sum of moments about a point different from the center of mass is not equal to zero, but equal to the moment of the resultant force, placed in the center of mass, about that point. This last equation can be also thought of as the comparison of the moments in free-body diagram with the moments in the following equivalent system:

Equivalent system in problem 5.9

Since the forces in this diagram and in the free-body diagram are equivalent, the total force and the total moment about any point must be the same the two diagrams.

Problem 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

(a) The external forces acting on the man are his weight, 891.8 N, the tension in the rope, , the normal force exerted by the ground on his foot, , and the static frictional force, , on his foot; the free-body diagram is the following

Forces on a man

The resultant couple in any point will be zero, since the man is in equilibrium. The resultant couple in pint A will not include the forces acting on the foot and will lead directly to the value of the tension in the rope:

and the tension in the rope is:

(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

The friction in the tires is static and the frictional forces are in the direction opposite to the motion of the truck, because none of the wheels have traction. The resistance of the air was neglected, since the speed must be low.

Problem 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

Problem 5.12, bloco

There are four forces acting on the body: its weight, , the normal force, , the tension in the string, , and the kinetic frictional force, ; the free-body diagram of the block is shown in the figure on the side. Since the acceleration does not have a vertical component, the sum of the vertical forces is zero; therefore, is equal to the weight, which is 49 N. Since the friction is kinetic, the frictional force is equal to N and the sum of the horizontal forces is:

Problema 5.12, roldana

There are four forces action on the pulley: its weight, the tensions on the two sides of the string, and , and a force in the axle. If the pulley's radius is , its moment of inertia around its axle is . Since the string does not slip over the pulley, the angular acceleration of the pulley is equal to , where is the acceleration of the block and the cylinder. The equation of motion for the pulley is then:

And substituting the expression previously obtained for we get,

Problema 5.12, cilindro

The only two forces acting in the cylinder are the string tension and the weight; the equation of motion is:

And substituting the expression previously obtained for the value of the acceleration can be found