Physics Help >> Index to Physics Homework Help » Friction, Weight, Elevators

Friction and weight are common experiences and taken for granted, but were not always understood. Aristotle believed that objects that contained earth fell because in Earth was at the center of the universe, and that was the natural order of things. We know know that weight, which is the force of gravity between objects, not only makes apples fall from trees, but keeps the Moon circling the Earth, and the Earth circling the sun. Aristotle also believed (and some students still believe) that the natural state of an object is to be at rest. That is, without a force pushing it, an object will slow down to a stop. This view overlooks friction, which is the force that causes most objects we see to come to rest. Now we know that there is nothing pushing planets along their orbit, and that without gravity to hold them in orbit, they would continue to move in a straight line until interfered with.

A. Weight, Vertical Forces, and Elevator Problems

Examples

1. What is the weight of a 200 kg object?

2. A woman exerts a force of 500 N straight up to lift a 35 kg basket. What is the acceleration of the basket?

Answers to Examples 1 and 2

1.

Near the Earth's surface, objects fall at 9.8 m/s/s. If air resistance is negligible then gravity is the only force acting on the falling object. Applying Newton's Second Law of Motion, a 200 kg object will have a weight of

F = ma = (200 kg)(9.8 m/s/s) = 1960 N

2. Free body diagram:

Sum the forces:
F_net = F_A + mg
F_net = 500 N + (35 kg)(-9.8 m/s²)
F_net = 157 N [up]

Second Law:
F_net = ma
F_net = (35.0 kg)a

Substitute:
157 N [up] = (35.0 kg)a
a = 4.49 m/s² [up]

Problems

(See section A answers.)

1. What is the acceleration of 2 falling sky divers (mass including parachute 132 kg) when the upward force is one-fourth their weight?

2. The instruments attached to a weather balloon have a mass of 6.9 kg. The balloon is released and exerts an upward force of 92.8 N on the instruments.

a. What is the acceleration of the balloon and the instruments?
b. After the balloon has accelerated for 10.7 s the instruments are released. What is the velocity of the instruments at the moment of their release?

3. An elevator is moving downward with an acceleration of 2.0 m/s². What is the force exerted on the elevator floor by a 75 kg person.

4. You are standing on a bathroom scale in an elevator and wish to calculate what the scale would read while experiencing the following scenarios: (just derive a working equation)
a)You are standing in the elevator that just began accelerating downwards as it approaches and comes to a stop at the 11th floor
b) You are standing in the elevator that just began accelerating upwards as it approaches and comes to a stop at the 10th floor.

5. Crates of mass 50.0 kg must be hoisted onto a platform 8.0 m above the ground. A person exerts 600.0 N of force on a rope that goes up and over a pulley suspended from the ceiling. The other end of the rope is attached to the 50.0 kg mass. How long will it take the person to lift a crate from the ground to the platform while exerting maximum force?

6. An elevator of mass 3000 kg is ascending at a steady speed of 2.0 m/s. What is the tension in the cable supporting the cable?

7. A person whose weight is 520 N is being pulled up vertically by a rope from the bottom of a cave that is 35.1 m deep. The maximum tension that the rope can withstand without breaking is 569 N. What is the shortest time, starting from rest, in which the person can be brought out of the cave?

8. Your groceries are in a bag with paper handles. The handles will tear off if a force greater than 51.0 N is applied to them. What is the greatest mass if groceries that can be lifted safely with this bag, given that the bag is raised (a) with constant speed, or (b) with an acceleration of 1.35 m/s squared? 

9. How much tension must a rope withstand if it is used to accelerate a 1300 kg car vertically upward at 0.90 m/s/s?

B. Horizontal and Friction Problems

Example

3. A 20.0 kg mass is pulled by along a surface by a horizontal force of 100 N. Friction is 20.0 N. What is the acceleration of the mass?

Answer to Example 3:

Free body diagram (whenever there are two or more forces):

Sum of forces (vertical forces cancel as evidenced by lack of acceleration in the vertical dimension)
F_net = T + F_f
F_net = (100 N) + (-20.0 N) = 80 N

Problems

(See section B answers.)

1. During takeoff, the four engines of a plane produce a thrust of 40 kN each. If the friction force equals 40 kN and the mass of the plane is 20,000 kg, calculate the time it will take the plane to reach a speed of 33m/s.

C. Coefficient of Friction Problems

Example

4.

A 49-N block is pulled by a horizontal force of 50.0 N along a rough horizontal surface at a constant acceleration of 6 m/s/s. What is the coefficient of friction?

Answer to example 4:

F_g = 49-N

T = 50.0 N

a = 6 m/s/s.

What is the coefficient of friction?

Free-body diagram:

Sum the forces:
In the vertical axes forces are equal and opposite as there is no vertical acceleration.

F_N = -F_g

F_N = mg = 49 N

Also, m = 49/9.81 = 4.99 kg

In the horizontal axis,

F_net = T + F_f

F_net = (50 N) + F_f

Newton's Second Law:
F_net = ma

F_net = (4.99 kg)(6) = 30.0 N

Problems

(See section C answers.)

1. A 15.0 kg block is pulled by a horizontal force of 30.0 N along a rough horizontal surface at constant velocity. What is the coefficient of friction?

2. A car is traveling at 60.6 mph on a horizontal highway. The acceleration of gravity is 9.8 m/s². If the coefficient of friction between road and tires on a rainy day is 0.056, what is the minimum distance in which the car will stop? Answer in units of m.

3. If the coefficient of kinetic friction between a 35 kg crate and the floor is .30, what horizontal force is required to move the crate at a steady speed across the floor? What horizontal force is required if µ_k is zero?

4. A 300 lbs car is loose inside a shopping mall where uk=0.5 If the car is initially traveling at 60 f/sec.
a) what is the kinetic friction force, if the brakes are locked?
b) what distance will it take to stop if the brakes are locked?

5. Car A is towing car B. Both cars have the same mass m_a = m_b = 1000 kg. Car A has four-wheel drive, and

the static coefficient of friction between its tires and the road is µ_s = 1.00 Neglect any friction force acting on

car B. Car A is accelerating at 1m/s/s. Calculate the tension in the horizontal rope connecting the cars.

6. A 4.0 kg toboggan rests on a frictionless icy surface and a 2.0 kg box rests on top of the toboggan. The coefficient of static friction between the box and toboggan is .60, whereas kinetic friction coefficient is .51.. The box is pulled by a 30 N force along the horizontal parallel to the icy floor. What are the magnitudes and directions of the resulting accelerations of the block and toboggan?

7. Four blocks are arranged on a frictionless surface. Two smaller placed on two larger, and a cord connects the two top (small) blocks, and a cord is used to pull the first large block. There is friction only between the top and bottom blocks. What is the maximum value of F applied to one of the blocks to make all four move with the same acceleration?

8. A block of mass 1.95 kg slides with an initial speed of 4.33 m/s on a smooth, horizontal surface. The block now encounters a rough patch with a coefficient of kinetic friction given by µ_k = 0.260. What is the acceleration of the block when it is in the rough patch?

9. A packing crate is placed on a plane inclined at an angle of 35° from horizontal. If the coefficient of static friction between the crate and the plane is 0.65, will the crate slide down the plane?

10. Suppose that you are standing on a train accelerating at 0.10g. What minimum coefficient of static friction must exist between your feet and the floor if you are not to slide?

11. In attempting to move a heavy couch (mass 200 kg) across a rug-covered floor, a man finds he must exert a horizontal force of 700 N to get the couch to barely move. Once the couch starts moving, the man continues to push with 700 N and his daughter, a physics major, estimates that it then accelerates at 1.10 m/s². Determine (a) the coefficient of static friction and (b) the coefficient of kinetic friction between the couch and the rug.

12. A 2500kg car traveling 14.0 m/s on an icy, level road approaches an intersection. The brakes lock and the car skids 25.0m. What is the coefficient of friction between the tires and the surface?

D. Problems Involving 2-Dimensional Forces

Example

5. A block is pulled by a string that makes an angle of 25º to the horizontal. If the mass of the block is 12.0 kg and the coefficient of friction is 0.25, what force would keep the block moving at a constant velocity?

Answer to Example 5:

Free-body diagram:

Sum the forces:
in the vertical

F_net = mg - F_N - T_v

Since there is no acceleration in the vertical axis, F_net = 0, and

F_N = mg - T_v = mg - Tsinø = (12 kg)(9.8) - Tsin25º

F_N = (118 N) - T(.4226)

In the horizontal

F_net = T_h - F_f

Since there is no acceleration in the horizontal axis, F_net = 0, and

T_h = F_f

T_h = Tcos25º = T(.9063)

F_f = µF_N = (0.25){(118 N) - T(.4226)} = 29.5 - T(0.1057)

Since T_h = F_f ,

T(.9063) = 29.5 - T(0.1057)

T(.8007) = 29.5

T = 36.8 N

Problems

(See section D answers.)

1. A clerk moves a box of cans down an aisle by pulling on a strap attached to the box. The clerk pulls with a force of 185N at an angle of 25.0 degrees to the horizontal. The box has a mass of 35.0 kg and the coefficient of kinetic friction between the box and floor is .450. Find the acceleration of the box.
2. A person pushes a 14.0 kg lawn mower at constant speed with a force of F = 88.0 N directed along the handle, which is at an angle of 45.0 degrees to the horizontal. Calculate

a. the horizontal friction force on the mower, then

b. the normal force exerted vertically upward on the mower by the ground.

3. A 4.1 kg box is pushed along a horizontal surface by a force F of magnitude 21 N at an angle of 35º with the horizontal. If the coefficient of kinetic friction between the box and the floor is .20. Calculate acceleration of the box.

4.

An electric winch is used to raise a 40-kg package and a 10-kg vertically up the side of a building. There is a frictional force of 60 N acting between the wall and the 40-kg package. The cable makes a 15º angle with the wall.
a) Calculate the force the winch must exert on the cable to slide the packages at a constant speed up the wall.

b) If the packages are lowed with an acceleration of 1.0m/s/s, what is the tension in the cable connecting the two packages?

5. You pull a 16.9 kg suitcase across the floor with a strap that is at an angle of 46.8° above the horizontal. Calculate the normal force on the suitcase, given that the suitcase moves with constant speed and that the coefficient of kinetic friction between the suitcase and the floor is 0.386. Also calculate the tension in the strap.

6. A 100kg block is pulled at a constant speed along a horizontal surface. The horizontal component of the force is 400N, while the vertical component of the force is 300N up. Find the kinetic friction, normal force and coefficient of friction.

7. A 100kg block is pushed at a constant speed along a horizontal surface. The horizontal component of the force is 400N, while the vertical component of the force is 300N down. Find the kinetic friction, normal force and coefficient of friction.

ANSWERS

For solutions to all the problems on this page click here.

C9.

We find the maximum angle the plane can be inclined without the crate sliding.

Since acceleration is zero, components along the y-axis cancel, and components along the x-axis cancel.

In other words, F_N = F_y and F_f = F_x .

Therefore, F_N = mgcosø and F_f = mgsinø

But F_f = µF_N , so

mgsinø = µmgcosø .

This yields µ = tanø .

Since µ = 0.65, ø = 33.0°

The crate will slide.

For solutions to all the problems on this page click here.

D7.

Since acceleration is zero, net force is zero. Therefore, F_f = F_x and kinetic friction is 400N.

Further, since net force is zero,

mg + F_y = N

(100)(9.81) + 300 = N

N = 1281N

Finally,

µ =F_f / N = 400/1281 = 0.312

For solutions to all the problems on this page click here.