NEWTON’S LESSON 12



NEWTON’S LESSON 12

INCLINE PLANES WITH FRICTION AND OTHER FORCES

In the presence of friction or other forces (applied force, tensional forces, etc.), the situation is slightly more complicated.

The perpendicular component of force still balances the normal force since objects do not accelerate perpendicular to the incline. The perpendicular component and the normal force add to 0 N.

Yet the frictional force must also be considered when determining the net force. The parallel component and the friction force add together to yield 5 N. The net force is 5 N, directed along the incline towards the floor.

EXAMPLE: This free-body diagram shows the forces acting upon a 100-kg crate which is sliding down an inclined plane. The plane is inclined at an angle of 30 degrees. The coefficient of friction between the crate and the incline is 0.3. Determine the net force and acceleration of the crate.

First, find the force of gravity acting upon the crate and the components of this force parallel and perpendicular to the incline.

Second, determine what you know:

The force of gravity is 980 N

The components of this force are Fgx = 490 N (980 N • sin 30 degrees) and Fgy = 849 N (980 N • cos30 degrees).

Now the normal force can be determined to be 849 N (it must balance the perpendicular component of the weight vector).

The force of friction can be determined from the value of the normal force and the coefficient of friction; Ff is 255 N (Ff = μ*FN= 0.3 • 849 N).

Determine the net force: The net force is the vector sum of all the forces. The forces directed perpendicular to the incline balance; the forces directed parallel to the incline do not balance.

The net force is Fnet = Fgx - Ff

235 N = (490 N - 255 N).

The acceleration is 2.35 m/s2 down the incline (a = Fnet/m = 235 N/100 kg).

 

Example:

[pic]

 

Diagram A

The Fg can be calculated from the mass of the object.

Fg = m • g = (1000 kg) • (9.8 m/s/s) = 9800 N

The parallel and perpendicular components of the gravity force can be determined from their respective equations:

Fgx = m • g • sin (45 degrees) = 6930 N

Fgy = m • g • cos (45 degrees) = 6930 N

The forces directed perpendicular to the incline balance each other. Thus FN is equal to Fy.

FN = 6930 N

There is no other force parallel to the incline to counteract the parallel component of gravity. Thus, the net force is equal to the Fgx value.

Fnet = 6930 N, down the incline

The acceleration is can be found from a = Fnet / m :

 a = Fnet / m = (6930 N) / (1000 kg)

a = 6.93 m/s2, down the incline

 

Diagram B

The Fg can be calculated from the mass of the object.

Fg = m • g = (1000 kg) • (10 m/s/s) = 9800 N

The parallel and perpendicular components of the gravity force can be determined from their respective equations:

Fgx = m • g • sin (60 degrees) = 8487 N

Fgy = m • g • cos (60 degrees) = 4900 N

The forces directed perpendicular to the incline balance each other. Thus FN is equal to Fy.

FN = 4900 N

There is no other force parallel to the incline to counteract the parallel component of gravity. Thus, the net force is equal to the Fgx value.

Fnet = 8487 N, down the incline

The acceleration is can be found from a = Fnet / m :

 a = Fnet / m = (8487 N) / (1000 kg)

a = 8.49 m/s2, down the incline

The affects of the incline angle on the acceleration of a roller coaster (or any object on an incline) can be observed in the two previous problems.

As the angle is increased, the acceleration of the object is increased.

As the angle increases, the component of force parallel to the incline increases and the component of force perpendicular to the incline decreases. It is the parallel component of the weight vector which causes the acceleration. Thus, accelerations are greater at greater angles of incline.

[pic]

 

Roller coasters produce two thrills associated with the initial drop down a steep incline.

The thrill of acceleration is produced by using large angles of incline on the first drop; such large angles increase the value of the parallel component of the weight vector (the component which causes acceleration).

The thrill of weightlessness is produced by reducing the magnitude of the normal force to values less than their usual values. It is important to recognize that the thrill of weightlessness is a feeling associated with a lower than usual normal force. Typically, a person weighing 700 N will experience a 700 N normal force when sitting in a chair. However, if the chair is accelerating down a 60-degrees incline, then the person will experience a 350 Newton normal force. This value is less than normal and contributes to the feeling of weighing less than one's normal weight - i.e., weightlessness.

 HOMEWORK NEWTON’S LESSON 11

[pic] 

• Read the problem carefully and diagram the physical situation (free-body diagram)

• Identify known and unknown information in an organized manner (write on diagram)

• Take the time to plot out a strategy prior to beginning the solution

• Identify an appropriate formula to use

• Perform step-by-step algebraic manipulations  

1. A student places two objects on a physics book and slowly tilts the cover. At a small

angle, object A starts to slide. At a large angle, object B starts to slide. Which object has the greater coefficient of static friction? Explain.

2. Lon Scaper is doing some lawn work when a 2.0-kg tire escapes from his wheelbarrow and begins rolling down a steep hill (a 30° incline) in San Francisco. Sketch the parallel and perpendicular components of this weight vector. Determine the magnitude of the components using trigonometric functions. Then determine the acceleration of the tire. Ignore resistance force.

 Finally, determine which one of the velocity-time graph would represent the motion of the tire as it rolls down the incline.

[pic]

Explain your answer.

3. In each of the following diagrams, a 100.0-kg box is sliding down a frictional surface at a constant speed of 0.200 m/s. The incline angle is different in each situation. Analyze each diagram and fill in the blanks.

Diagram A

[pic]

  

Diagram B

[pic]

4. A 30.0 kg object is on an inclined plane set at 40.0° to the horizontal. What is the acceleration experienced by the object if the coefficient of friction is 0.250?

5. Consider a 10.0 kg box on an inclined plane of 30o to the horizontal having a coefficient of friction of μ= 0.50.

1. What is the force of friction on the box?

2. What is the acceleration of the box?

6. Suppose that you now want to drag a 5.0-kg mass up and down a frictionless 15º inclined plane.

 

[pic]

1. How much force must you apply to a string acting parallel to the incline's surface to slide the 5.0-kg mass up or down the incline at a constant velocity (no acceleration)?

2. How much upward force would be needed to accelerate the 5.0-kg mass up this incline at 3.0 m/sec2?

3. How much upward force would be needed to restrict the 5.0-kg mass' downward acceleration to 1.0 m/sec2?

4. If the 5.0-kg mass were allowed to slide down this incline without any additional applied forces acting upon it, what would be its acceleration down the incline?

7. A 10.0 kg box accelerates at 2.0 m/s2 as it slides down a ramp that makes an angle of 30 degrees with the horizontal. Find the coefficient of friction.

8. A city planner is working on the redesign of a hilly portion of a city. An important consideration is how steep the roads can be so that even low-powered cars can get up the hills without slowing down. It is given that a particular small car, with a mass of 1100kg, can accelerate on a level road from rest to 21 m/s in 14.0 s. Using this data, calculate the maximum steepness of the hill.

HINT: The direction of motion depends not only upon the applied force, the weight of the block, and also on the direction of the force of friction.

[pic]

9. An object is being pulled up a 15° incline against a frictional coefficient of 0.15, and requires a force of 835 N parallel to the surface of the ramp to move it at a constant speed. What is the weight of the object?

HOMEWORK KEY INCLINE PLANE PROBLEMS WITH FRICTION AND APPLIED FORCE

1. Object B

2. Graph B

3. A: Fg = 980 N, Ff = 490 N, FN = 849 N B: Fg = 980 N, Ff = 693 N, FN = 693 N

4. 4.43 m/s2 down the incline

5. Ff = 42 N up incline, a = 0.66 m/s2 down incline

6. 1.) 13 N up incline, 2.) 28 N up incline, 3.) 7.7 N up incline, 4.) 2.5 m/s2 down incline

7. 0.34

8. 8.8 degrees

9. 2.1 x 103 N

 

HOMEWORK NEWTON’S LESSON 12

1. A student places two objects on a physics book and slowly tilts the cover. At a small

angle, object A starts to slide. At a large angle, object B starts to slide. Which object has the greater coefficient of static friction? Explain.

Answer: Object B has a greater coefficient of static friction. As the incline increases, the force of gravity in the x plane (parallel to surface of incline) increases. This parallel force is in the opposite direction as the static friction force. As the incline increased, the force of gravity (parallel to the incline) surpassed the static coefficient of friction resulting in a downward acceleration. Therefore Object B has the greatest coefficient of static friction as it required the largest incline angle.

2. Lon Scaper is doing some lawn work when a 2.0-kg tire escapes from his wheelbarrow and begins rolling down a steep hill (a 30° incline) in San Francisco. Sketch the parallel and perpendicular components of this weight vector. Determine the magnitude of the components using trigonometric functions. Then determine the acceleration of the tire. Ignore resistance force.

 Finally, determine which one of the velocity-time graph would represent the motion of the tire as it rolls down the incline.

[pic]

Explain your answer.

 Answer: The Fg can be calculated from the mass of the object.

Fg = m • g = (2 kg) • (9.8 m/s/s) = 20 N

The parallel and perpendicular components of the gravity force can be determined from their respective equations:

Fgx = 19.6 N  • sin (30 degrees) = 9.8 N

Fgy = 19.6 N  • cos (30 degrees) = 17 N

The acceleration of the tire is the ratio of net force to mass:

a = Fnet / m = (9.8 N) / (2.00 kg) = 4.9 m/s2

The correct graph is Graph B since the tire will accelerate (speed up) down the hill.

3. In each of the following diagrams, a 100.0-kg box is sliding down a frictional surface at a constant speed of 0.200 m/s. The incline angle is different in each situation. Analyze each diagram and fill in the blanks.

Diagram A

[pic]

 

  

Diagram B

[pic]

 

 Answer:

Diagram A: The Fg can be calculated from the mass of the object.

Fg = m • g = (100 kg) • (9.8 m/s/s) = 980 N

The parallel and perpendicular components of the gravity force can be determined from their respective equations:

Fgx = m • g • sin (30 degrees) = 490 N

Fgy = m • g • cos (30 degrees) = 849 N

For a constant speed motion, the forces parallel to the incline must balance each other. Thus, the Ff is equal to the Fx.

Ff = 490 N

The forces directed perpendicular to the incline balance each other. Thus FN is equal to Fy.

FN = 849 N

Diagram B: The Fg can be calculated from the mass of the object.

Fg = m • g = (100 kg) • (9.8 m/s/s) = 980 N

The parallel and perpendicular components of the gravity force can be determined from their respective equations:

Fgx = m • g • sin (45 degrees) = 693 N

Fgx = m • g • cos (45 degrees) = 693 N

For a constant speed motion, the forces parallel to the incline must balance each other. Thus, the Ff is equal to the Fx.

Ff = 693 N

The forces directed perpendicular to the incline balance each other. Thus FN is equal to Fy.

FN = 693 N

 

4. A 30.0 kg object is on an inclined plane set at 40.0° to the horizontal. What is the acceleration experienced by the object if the coefficient of friction is 0.250?

Answer:

[pic]

5. Consider a 10.0 kg box on an inclined plane of 30o to the horizontal having a coefficient of friction of μ= 0.50.

1. What is the force of friction on the box?

2. What is the acceleration of the box?

Answer:

a. First, Draw a free body diagram

[pic]

The normal force acting on the box is mgcos30= 84.87 N, exactly the same as for the frictionless inclined plane. The force of friction is defined as Ff = μFN, so plugging in the appropriate values for [pic]and FN:

Ff = μFN = 0.500*84.87 = 42.435 N = 42 N, upward.

b. The net force acting on the box is the difference between the downhill gravitational force and the force of friction. From this balanced force equation you can determine acceleration:

Fnet = mgsin30 - Ff

ma = mgsin30 - Ff

10.0*a = 10.0(9.8)(0.500) – 42.435

a = 0.6565 = 0.66 m/s2 downward

Because [pic], the direction of the acceleration is in the downhill direction.

6. Suppose that you now want to drag a 5.0-kg mass up and down a frictionless 15º inclined plane.

 

[pic]

1. How much force must you apply to a string acting parallel to the incline's surface to slide the 5.0-kg mass up or down the incline at a constant velocity (no acceleration)?

2. How much upward force would be needed to accelerate the 5.0-kg mass up this incline at 3.0 m/sec2?

3. How much upward force would be needed to restrict the 5.0-kg mass' downward acceleration to 1.0 m/sec2?

4. If the 5.0-kg mass were allowed to slide down this incline without any additional applied forces acting upon it, what would be its acceleration down the incline?

Answer:

1. The parallel components of the gravity force:

Fgx = m • g • sin (45 degrees) = 5.0*9.8*sin (15 degrees) = 12.682 N

For a constant speed motion, the forces parallel to the incline must balance each other. Fnet = Fgx - Fapp = 0. Thus, the Fapp is equal to the Fx.

Fapp = 12.682 N = 13 N upwards

2. If the object is accelerating up the incline at 3.0 m/s2 then the Fnet = ma = 5.0*3.0 = 15 N up the incline.

Determine an equation the represents the forces acting on the object in the x plane.

Fnet = Fapp - Fgx

15 N = Fapp – 12.682

Fapp = 27.682 N = 28 N, up the incline

3. If the object is accelerating down the incline at 1.0 m/s2 then the Fnet = ma = 5.0*(1.0) = 5.0 N

Determine an equation the represents the forces acting on the object in the x plane.

Fnet = Fgx - Fapp

5.0 N =12.682 - Fapp

Fapp = 7.682 N = 7.7 N, up the incline

4. If no applied force, the object would accelerate down the incline due to the parallel force of gravity alone:

Fgx = Fnet

12.682 = 5.0(a)

a = 2.536 = 2.5 m/s2 down incline

7. A 10.0 kg box accelerates at 2.0 m/s2 as it slides down a ramp that makes an angle of 30 degrees with the horizontal. Find the coefficient of friction.

The Fg can be calculated from the mass of the object.

Fg = m • g = (10 kg) • (9.8 m/s/s) = 98 N

The parallel and perpendicular components of the gravity force can be determined from their respective equations:

Fgx = m • g • sin (30 degrees) = 49 N

Fgy = m • g • cos (30 degrees) = 84.9 N

For a box that accelerates at 2.00 m/s2, the Fnet = ma = 10.0*2.00 = 20.0 N

Fnet = Fgx - Ff

20.0 N = 49 - Ff

Ff = 29 N

The forces directed perpendicular to the incline balance each other. Thus FN is equal to Fy.

FN = 84.9 N

Ff = μFN

29 = μ84.9

μ ’ 0.34

8. A city planner is working on the redesign of a hilly portion of a city. An important consideration is how steep the roads can be so that even low-powered cars can get up the hills without slowing down. It is given that a particular small car, with a mass of 1100kg, can accelerate on a level road from rest to 21 m/s in 14.0 s. Using this data, calculate the maximum steepness of the hill.

Answer:

First, think about the question, what is going on, and what are the forces acting on the car?

Second, draw a free body diagram.

Third, determine the acceleration on the object:

vf = vi + at

21 = 0 + a(14)

a = 1.5 m/s up the hill.

This is the rate that a car can accelerate at. The Fapp on the object up the hill will counteract the Fgx down the hill resulting in a Fnet of zero overall. If these two are equal then the car will not slow down (accelerate) but rather maintain a constant velocity up the hill.

Fapp = Fgx

ma= mgsinθ

1.5 = 9.8 sin θ

θ = 8.8 degrees

HINT: The direction of motion depends not only upon the applied force, the weight of the block, and also on the direction of the force of friction.

[pic]

9. An object is being pulled up a 15° incline against a frictional coefficient of 0.15, and requires a force of 835 N parallel to the surface of the ramp to move it at a constant speed. What is the weight of the object?

 First, draw a free body diagram.

Second, since an Fapp of 835 N is required to move the object at a constant speed (no acceleration), the Fnet must be zero and Fgx + Ff must add to Fapp:

0 = Fnet = Fapp - Fgx - Ff

Fapp = Fgx + Ff

835 N = mgsin 15 + mgcos 15(0.15) recall: Ff = FN μ = Fgy* μ= mgcos 15*μ

FN = Fgy

835 = mg(sin 15 + (cos 15)*0.15)

mg = 835/(0.2588 + 0.144888)

mg = Fg = 2068.42 = 2.1 x 103 N

The weight of the object is 2.1 x 103 N.

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