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Friction and weight are common experiences and taken for granted, but were not always understood. Aristotle believed that objects that contained the "element," earth, fell because Earth was at the center of the universe, and that was the natural order of things. We now 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. Without gravity to hold them in orbit, they would continue to move in a straight line until interfered with.

Aristotle also believed (and some students still believe) that the natural state of an object is to be at rest. That is, he believed unless a force was pushing an object, the object will slow down to a stop. This view overlooks **friction**, the force that resists the movement of surfaces sliding against each other. Friction causes most objects we see to come to rest. Unlike Aristotle, we know that there is nothing pushing planets along their orbit: since there is no friction to slow down planets, they will continue to move until interfered with.

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} + m*g*

**F**_{net} = 500 N + (35 kg)(-9.8 m/s^{2})

**F**_{net} = 157 N [up]

Second Law:

**F**_{net} = m**a**

**F**_{net} = (35.0 kg)**a**

Substitute:

157 N [up] = (35.0 kg)**a**

**a** = 4.49 m/s^{2} [up]

(See below for answers)

1. What is the acceleration of a falling crate (mass including parachute 132 kg) when the upward force or air resistance is one-fourth its weight?

2. The payload lifted by a balloon has a mass of 6.9 kg. The balloon exerts an upward force of 92.8 N on the payload.

a. What is the acceleration of the balloon and the payload?

b. The balloon accelerates from rest for 10.7 s, at which point the payload is released. What is the velocity of the payload at the moment of its release?

3. An elevator is moving downward with an acceleration of 2.0 m/s^{2}. 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. A constant force of 600 N is applied straight up on a 50 kg crate. How long will it take for the crate to rise 8.0 m from rest?

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. An 80-kg stuntman steps out of a window 30 meters above layers of mattresses. Air resistance exerts 100-N on the stuntman. What is his velocity of impact?

8. The string used to lift a pail of water will break if there is more than 51.0 N of tension. What is the greatest combined mass of the pail and water if they are lifted (a) with constant speed? (b) with an acceleration of 1.35 m/s/s?

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

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

(See below for answers)

1. A jet engine generates 160 kN of force as it propels a 20,000kg plane down a runway. If 40 kN of friction opposes the plane, how much time it will take the plane to reach a speed of 33m/s from rest?

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.

First, draw a 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} = m**a**

**F**_{net} = (4.99 kg)(6) = 30.0 N

(See below for 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/s^{2}. 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 lb golf cart is traveling at 60 f/sec on a surface with a coefficient of kinetic friction equal to 0.5.

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. Two large blocks, each with mass = M, are placed on a frictionless surface. A smaller block, with mass = m, is placed on each of the two larger, with a cord connecting the two top (small) blocks. Another cord is used to pull the first large block. The coefficient of friction between the top and bottom blocks is µ. What is the maximum force, F, that can be applied to the first large block 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 carpet, 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/s^{2}. Determine (a) the coefficient of static friction and (b) the coefficient of kinetic friction between the couch and the carpet.

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?

13. A 10,000-kg load sits unsecured on the flat bed of a 20,000-kg truck moving at 12.0 m/sec. The load has a coefficient of static friction of 0.500 with the truck bed. Calculate the minimum stopping distance for which the load will not slide forward relative to the truck.

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:

First, draw a free-body diagram.

Sum the forces in the vertical axis:

**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 axis:

**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(1.012) = 29.5

T = 29.2 N

(See below for answers)

1. A rope attached to a 35.0 kg box makes an angle of 25.0 degrees with the horizontal. A force of 185N is applied to the rope, 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. A 40-kg package and a 10-kg package slide vertically up a wall. The angle between the cable and the wall is 15º. The friction between the wall and the 40-kg package is 60 N.

a. What is the tension in the upper cable if the packages move at a constant speed?

b. What is the tension in the cable connecting the packages if they are lowered at 1.0 m/s/s?

5. A strap attached to a 16.9 kg chest makes an angle of 46.8° above the horizontal. The coefficient of kinetic friction between the chest and the floor is 0.386. Calculate the tension in the strap and the normal force on the chest when the chest moves with constant speed.

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.

8. An 18-kg suitcase is pulled at constant speed by a 43-N force applied to an attached strap.. Friction is 27 N. a. What angle does the strap make with the horizontal? b. What is the normal force exerted on the suitcase?

Selected solutions are printed below.

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

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