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UNIT IV: Worksheet 3

For each of the problems below, carefully draw a force diagram of the system before attempting to solve the problem.

1. Determine the tension in each cable in case A and case B.

Case A Case B

2. Determine tension in each cable. (Hint: There is more than one way to define the system.)

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3. The object hung from the cable has a weight of 25 N. Write the equation for the sum of the forces in the y-direction. What is the tension in the cable?

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Repeat the problem above with a 5° angle. How does the tension compare?

4. The cable at left exerts a -30 N force.

a. Write the equation for the sum of the forces in the x-direction. What is the value of T2?

b. Write the equation for the sum of the forces in the y-direction. What is the force of gravity acting on the ball?

5. The box on the frictionless ramp is held at rest by the tension force. The mass of the box is

20 kg. What is the value of the tension force?

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What is the value of the normal force?

6. In the system below the pulley and ramp are frictionless and the block is in static equilibrium. What is the mass of the block on the ramp?

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7. A man pulls a 50 kg box at constant speed across the floor. He applies a 200 N force at an angle of 30°.

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a. Sum the forces in the x-direction. What is the value of the frictional force opposing the motion?

b. Sum the forces in the y-direction. What is the value of the normal force?

8. A man pushes a 2.0 kg broom at constant speed across the floor. The broom handle makes a 50° angle with the floor. He pushes the broom with a 5.0 N force.

a. Sum the forces in the y-direction. What is the value of the normal force?

b. Sum of the forces in the x-direction. What is the value of the frictional force opposing the motion?

c. If the frictional force were suddenly reduced to zero, what would happen to the broom?

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T1

Case A: FT = 5kg * 10m/s2 = 50N

Case B: Two strings mean 50N is shared equally. Each has 25N of tension.

5 kg

30(

T2

T

Top: FT = 11kg * 10m/s2 = 110N

Bottom: FT = 4kg * 10m/s2 = 40N

Both cables have the same tension.

y-direction: FTy = ½ * 25N = 12.5N

FT = 12.5N / sin(30°) = 25N.

FT = 12.5N / sin(5°) = 143N.

x-direction: FT2x = -(FT1) = 30N

FT2 = 30N / sin(30°) = 60N.

FT2y = 30N / tan(30°) = 51.2 N.

Fg = -51.2N

- Fgy = FN = (cos 30°)*200N = 173.2N

Fgx = - FT = sin(30°)*200N = 100N.

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- Fgy = FN = (cos 30°)*200N = 173.2N

FT = -Fg of hanging mass = 200N = Fgx

Fg = 200N / sin 35°= 348.7N

Mass = 34.87 kg

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x-direction:

FTx = -(Ff) = cos 30°(200N) = 173.2N

FTy + Fn = -Fg

sin(30°)*200N + Fn = 500N.

Fn = 400N

y-direction:

Fpy + Fg = -(Fn)

sin 50°(5.0N) + 20N = 23.8N = Fn

x-direction:

FTx = -Ff

cos(50°)*50N = 3.21N.

Fn = 400N

x-direction:

FTx = -Ff

cos(50°)*50N = 3.21N.

The 3.21 N force would accelerate the 2kg broom:

a = F/m = 3.21N/2.0kg = 1.61m/s2

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