In Fig. 6-25, blocks A and B have weights of 49 N and 21 N, respectively.
(a) Determine the minimum weight of block C to keep A from sliding, if µs between A and the table is 0.20.
(b) Block C suddenly is lifted off A. What is the acceleration of block A, if µk between A and the table is 0.15?
I'm not really sure where to even start. I've summed my forces, I just don't know how I go about finding the Block C's mass.
Never mind. I realized that you want the force of friction to equal the force of tension. Thus when you set it up, you find the normal force and subtract the current way.
Ft=Ff
Ft=mu x N
N-Ma=Mc
To solve this problem, you need to consider the forces acting on the blocks and use Newton's laws of motion. Let's break down the steps:
(a) To determine the minimum weight of block C to keep block A from sliding, we need to analyze the forces acting on block A.
1. Identify the forces acting on block A:
- Weight of block A: W_A = 49 N (acts vertically downward)
- Normal force exerted by the table on block A: N_A (acts vertically upward)
- Frictional force between block A and the table: F_friction = μs * N_A (acts horizontally)
- Force exerted by block C on block A: F_C (acts horizontally)
2. Apply Newton's second law to block A along the horizontal direction:
∑F_horizontal = F_C - F_friction = ma
Substitute the expression for frictional force: F_C - μs * N_A = ma
3. Sum the forces acting vertically for block A:
∑F_vertical = N_A - W_A = 0
N_A - 49 N = 0
N_A = 49 N
4. Substitute N_A into the equation from step 2:
F_C - μs * 49 N = ma
5. Since we want to find the minimum weight of block C to prevent sliding, we can assume that the acceleration is zero. Therefore, a = 0.
6. Solve the equation for F_C:
F_C - μs * 49 N = 0
F_C = μs * 49 N
F_C = 0.20 * 49 N
F_C = 9.8 N
Therefore, the minimum weight of block C required to keep block A from sliding is 9.8 N.
(b) Now, let's calculate the acceleration of block A when block C is suddenly lifted off.
1. Identify the forces acting on block A:
- Weight of block A: W_A = 49 N
- Normal force exerted by the table on block A: N_A = 49 N
- Frictional force between block A and the table: F_friction = μk * N_A (acts horizontally)
- No force exerted by block C because it is lifted off
2. Apply Newton's second law to block A along the horizontal direction:
∑F_horizontal = F_friction = ma
Substitute the expression for frictional force: μk * 49 N = ma
3. Solve the equation for acceleration:
μk * 49 N = ma
a = (μk * 49 N) / m
To calculate the acceleration, you will need to know the mass of block A, which is not provided in the question. The acceleration depends on the mass of block A and the coefficient of kinetic friction (μk). With this information, you can substitute the appropriate values to find the acceleration.