Block A of mass 2 kg is on block B of mass 6 kg as shown in the above figure. The lower block is on a frictionless surface while the coefficient of static friction is 0.25 between the two blocks.
a) If they are moving at constant velocity, what is the frictional force between A and B?
b) What is the maximum horizontal force that can be applied to B without A slipping?
a) If the blocks are moving at constant velocity, it means that the net force on the system is zero. Since there is no friction on the lower block, the only force acting on the blocks is the force of friction between A and B. According to Newton's second law, the force of friction is equal to the product of the coefficient of static friction and the normal force:
Frictional force = μ * Normal force
The normal force on block A is equal to its weight, which is the mass of A multiplied by the acceleration due to gravity:
Normal force on block A = m_A * g = 2 kg * 9.8 m/s^2 = 19.6 N
The normal force on block B is equal to the weight of block A plus the weight of block B:
Normal force on block B = (m_A + m_B) * g = (2 kg + 6 kg) * 9.8 m/s^2 = 78.4 N
Thus, the frictional force between A and B is:
Frictional force = 0.25 * 78.4 N = 19.6 N
b) The maximum horizontal force that can be applied to block B without block A slipping is equal to the product of the coefficient of static friction and the normal force on block A:
Max horizontal force on block B = μ * Normal force on block A
Using the same values as before:
Max horizontal force on block B = 0.25 * 19.6 N = 4.9 N