Solved Problems

4. Vector dynamics

Problem 1

A 70 kg person climbs to the sixth floor of a building using an elevator. The elevator starts from rest on the ground floor, accelerates until the second floor with uniform acceleration of 2 m/s2, keeps the speed constant between the second and fourth floors and slows down between the fourth and sixth floors with uniform acceleration of −2 m/s2. Determine the magnitude of the normal force in the person's feet in each part of the climb.


Problem 4.1

The figure on the right shows free-body diagram for the person, where is the normal force exerted by the elevator's floor on the person's feet. The sum of the forces component's in the direction of motion is

The tangential acceleration is positive between the ground floor and the second floor, and Newton's second law leads to:

Between the second and fourth floors, the resulting force is zero:

Finally, between the fourth and sixth floors, the tangential acceleration is negative:


Problem 5

A 72 kg man pushes an 8 kg wooden box on a horizontal floor, exerting a horizontal force on it that makes it slide on the floor. On the box is placed a book with 0.6 kg. The man, the box and the book move together, with acceleration equal to 0.5 m/s2. Determine the values ​​of the frictional forces between the floor and the box, between the box and the book and between the floor and the feet of the man, ignoring the air resistance and knowing that the coefficients of static friction ( ) and kinetic friction ( ) are: between the floor and the box, and ; between the box and the book, and ; between the ground and the feet of the man, and .

There are four points of contact between rigid bodies:

  1. Between the base of the book and the top of the box.
  2. Between the base of the box and the floor.
  3. Between the feet of the man and the floor.
  4. Between the man's hands and the side wall of the box where the man is pushing.

In point 1 there is a normal force, , vertical, and a horizontal static friction force, . There is static friction because the book is not sliding over the box. In point 2 there is a normal force, , vertical, and a horizontal frictional force, , which is kinetic friction because the box slides on the floor. In point 3 there is a normal force, , vertical, and a horizontal frictional force, , which is static friction because the man's feet do not slide on the floor (if they did, the box would not accelerate). In point 4 there is only a normal force, , because the man pushes the box horizontally.

The following figure shows the three free-body diagrams for the book, with mass , the box, with mass , and the man, with mass .

Man pushing a box

The normal force makes the box accelerate. The force that makes the book follow the motion of the box, with the same acceleration, is the frictional force , which acts in the same direction of and the force responsible for the acceleration of the man is the frictional force on his feet, also in the same direction. The frictional force is in the opposite direction of the acceleration and, since it is kinetic friction, its magnitude equals . The weights of the 3 bodies are: N, N and N.

The sum of the vertical and horizontal components of the forces on the book are (in SI units):

The sum of the components of the forces on the box are:

And the sums of the forces over the man are:

The maximum possible value of the frictional force is and the maximum possible value of Is . As the results obtained do not exceed these maximum values, the results are valid and the answer is: the frictional force between the box and the book is 0.3 N, the frictional force between the box and the floor is 0.2 × 84.28 = 16.856 N And the frictional force between the floor and the man's feet is 57.156 N.

Problem 6

A 1230 kg car climbs an 8 percent slope with constant speed. Determine:

  1. The value of the total frictional force (sum of the forces on the four tires).
  2. The minimum value of the coefficient of static friction between the road and the tires so that the car can climb up the ramp.
Car climbing a ramp

Problem 4.6

The figure shows the free-body diagram of the car, where and Are the sum of the normal forces and the friction forces in the four tires. Since the frictional force must point in the direction that the car is moving, otherwise the car would move down the road due to gravity, there must be static friction (at least on some of the tires). And since the velocity is constant, the acceleration is zero and the sum of external forces must also be zero. Using the system of axes shown in the figure, the sums of the and components of the forces are

(a) Since the angle is equal to the inclination of the road, then the second equation leads to

(b) The normal force is determined from the equation for the sum of the components of the forces

And since,

we then find that

Sum of forces in problem 4.6

This problem could have also been solved by placing the three forces one after each other, as shown in the figure on the right. Since the forces add up to zero, the three vectors form a triangle which in this case is rectangle and similar to the triangle that defines the inclination of the road. By similarity of triangles, we conclude that the frictional force is equal to , the normal reaction is equal to and the minimum coefficient of friction is then 8/100.


Problem 8

A sphere of radius and density falls freely inside a fluid with a density and viscosity coefficient . (a) Find the expressions for the terminal velocity when the resistance of the fluid is proportional to the velocity and when it is proportional to the square of the velocity. (b) Compute the terminal velocities of a sphere of diameter 1 cm made of steel, with density 7800 kg/m3, inside glycerin, water and air. In each case determine the value of the Reynolds number. Use the data in the following table:

FluidViscosity (kg/(m·s))Density (kg/m3)
Glycerin1.51200
Water10−31000
Air 1.8×10−51.2

Problem 4.8

(a) The figure shows the free-body diagram of the sphere, where stands for the mass of the sphere minus the mass of the fluid displaced by it and is the force of resistance exerted by the fluid. That force is initially zero, but it increases as the speed of the sphere increases. At the instant when the magnitude of that force equals the effective weight , the acceleration becomes zero and the sphere reaches a constant limit velocity at which the resistance force remains constant too. Therefore, to find the terminal velocity we solve the condition

In the case of the force of resistance proportional to the velocity, equation 4.12 for a sphere leads to the following expression

And in the case of the resistance force proportional to the square of the velocity, equation 4.14 for a sphere leads to the following expression

(b) In glycerin, as the viscosity is high, we initially assume that the resistance is proportional to the velocity, so the terminal velocity of the sphere is

Using the radius of the sphere, the Reynolds number is

That value, less than 1, confirms that the resistance force is in fact proportional to the velocity. The same calculations for water lead to the following results

Which shows that the result is inconsistent because the Reynolds number is very high. This means we must repeat the calculations assuming that the resistance force is proportional to the square of the velocity

Which is now a valid result because the Reynolds number is consistent with the expression used for the resistance force. Repeating the same calculations for the case of air, we find


Problem 10

To measure the coefficient of static friction between a block and a disc, the disc was rotated with constant angular acceleration rad/s2. The disc starts from rest at and at  s the block begins to skid over the disk. Determine the value of the coefficient of static friction.


Block on a disk

Problem 4.10

The figure shows the free-body diagram of the block, where Is the normal force and is the static friction force. Since there is no vertical movement, the normal force is equal to the weight. And the only force in the direction of the acceleration, the frictional force, must be equal to . Since the block initially shares the circular motion of the disc, it has the acceleration of a non-uniform circular motion, namely

As soon as the block begins to skid, the static friction force has its maximum value , so the static friction coefficient is:

To find the angular velocity at the instant the block begins to skid, we integrate the differential equation that defines the angular acceleration as the derivative of the angular velocity with respect to time

And replacing it in the expression for the coefficient of friction we get:

x