In Fig. 6-34, blocks A and B have weights of 46 N and 23 N, respectively. (a) Determine the minimum weight of block C to keep A from sliding if μs between A and the table is 0.21. (b) Block C suddenly is lifted off A. What is the acceleration of block A if μk between A and the table is 0.13?
To solve this problem, we need to consider the forces acting on each block and apply the principles of equilibrium and kinematics.
(a) To determine the minimum weight of block C to keep A from sliding, we need to find the maximum possible friction force that can prevent A from sliding. The formula for the maximum static friction force is given by:
f_s = μ_s * N,
where f_s is the maximum static friction force, μ_s is the coefficient of static friction, and N is the normal force.
In this case, the normal force acting on block A is equal to its weight, N_A = weight_A = 46 N.
Thus, the maximum static friction force between block A and the table is:
f_s = μ_s * N_A = 0.21 * 46 N ≈ 9.66 N.
Since the weight of block B is 23 N, the force exerted by block B on block A is its weight, W_B = 23 N, and it acts vertically downwards. To counteract this force and prevent block A from sliding, block C must exert an equal and opposite force vertically upwards.
Therefore, the minimum weight of block C to keep A from sliding is:
weight_C = W_B = 23 N.
(b) When block C is suddenly lifted off block A, the friction force between block A and the table changes from static friction to kinetic friction. The kinetic friction force is given by:
f_k = μ_k * N,
where f_k is the kinetic friction force, μ_k is the coefficient of kinetic friction, and N is the normal force.
Again, the normal force acting on block A is its weight, N_A = weight_A = 46 N.
So, the kinetic friction force between block A and the table is:
f_k = μ_k * N_A = 0.13 * 46 N ≈ 5.98 N.
Since the weight of block B is 23 N, the net force acting on block A is:
F_net = W_B - f_k = 23 N - 5.98 N ≈ 17.02 N (in the direction of block B).
The acceleration of block A can be found using Newton's second law:
F_net = m_A * a,
where F_net is the net force on block A, m_A is the mass of block A, and a is its acceleration.
Since the mass of block A is given by:
m_A = weight_A / g,
where g is the acceleration due to gravity (approximately 9.8 m/s^2),
the acceleration of block A is:
a = F_net / m_A = (17.02 N) / (weight_A / g) = (17.02 N) / (46 N / 9.8 m/s^2) ≈ 4.32 m/s^2.
Therefore, the acceleration of block A when block C is lifted off is approximately 4.32 m/s^2.