3. The 240-lb body in Fig. P-1078 is supported by wheels at B which roll freely without friction and by a skid at A under which the coefficient of friction is 0.40. Compute the value of P to cause an acceleration of (1/3)g.See answer

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3 The 240 lb body in Fig P 1078 is supported by wheels at B which roll freely without friction and by a skid at A under which the coefficient of friction is 040 Compute the value of P to cause an…

Question

3. The 240-lb body in Fig. P-1078 is supported by wheels at B which roll freely without friction and by a skid at A under which the coefficient of friction is 0.40. Compute the value of P to cause an acceleration of (1/3)g.

Basic Answer

Ideas for solving the problem：

This problem can be solved by applying Newton’s second law of motion and considering the forces and friction involved. We’ll use free body diagrams and equations of equilibrium.

Calculation step：

Step 1: Free Body Diagram
Draw a free body diagram of the body. The forces acting on the body are:

• Weight (W) = 240 lb, acting vertically downwards.
• Normal force at A (N_A), acting vertically upwards.
• Normal force at B (N_B), acting vertically upwards.
• Friction force at A (F_A) = μN_A = 0.40N_A, acting horizontally to oppose motion (to the left if P is to the right).
• Applied force P, acting horizontally.

Step 2: Equations of Motion
Apply Newton’s second law in the horizontal and vertical directions:

• ΣF_x = ma_x => P – F_A = ma_x
• ΣF_y = ma_y => N_A + N_B – W = 0 (Since there’s no vertical acceleration)

We are given that a_x = (1/3)g = (1/3)(32.2 ft/s²) ≈ 10.73 ft/s². The mass (m) can be calculated from the weight: m = W/g = 240 lb / 32.2 ft/s² ≈ 7.45 lb·s²/ft.

Step 3: Moment Equilibrium
Take moments about point B. This eliminates N_B from the equation. Assume counter-clockwise moments are positive:
ΣM_B = 0 => N_A(6 ft) – W(3 ft) + P(2 ft) = 0

Step 4: Solving the Equations
We have three equations and three unknowns (N_A, N_B, P).

1. P – 0.40N_A = (7.45 lb·s²/ft)(10.73 ft/s²) ≈ 79.9 lb
2. N_A + N_B – 240 lb = 0
3. 6N_A – 720 lb + 2P = 0

Substitute equation (1) into equation (3):
6N_A – 720 lb + 2(79.9 lb + 0.40N_A) = 0
6N_A – 720 lb + 159.8 lb + 0.80N_A = 0
6.8N_A = 560.2 lb
N_A ≈ 82.4 lb

Substitute N_A back into equation (1):
P – 0.40(82.4 lb) = 79.9 lb
P = 79.9 lb + 32.96 lb
P ≈ 112.86 lb

Step 5: Check for impending motion
We assumed that the body is accelerating to the right. If P were smaller, the body might not move. The solution is valid as long as the friction force at A is less than or equal to the maximum possible friction force (μN_A). In this case, F_A = 0.40 * 82.4 lb = 32.96 lb, which is less than the maximum possible friction force.

Final Answer:

The value of P required to cause an acceleration of (1/3)g is approximately 112.86 lb.

Highlights:

• Draw a clear free body diagram.
• Correctly apply Newton’s second law in both horizontal and vertical directions.
• Use moment equilibrium to eliminate one unknown.
• Solve the system of equations simultaneously.
• Check the validity of the assumptions made (impending motion). The friction force should be less than or equal to the maximum static friction.