Question

You are lowering two boxes, one on top of the other,

down a ramp by pulling on a rope parallel to the surface of the
ramp (Fig. E5.33). Both boxes move together at a constant speed
of 15.0 cm>s. The coefficient of kinetic friction between the ramp
and the lower box is 0.444, and the coefficient of static friction
between the two boxes is 0.800. (a) What force do you need to
exert to accomplish this? (b) What are the magnitude and direction of the friction force on the upper box?

Answer 1

To solve this problem, we need to use Newton's second law to find the net force acting on the two boxes. The net force is equal to the mass of the two boxes times the acceleration. Since the boxes are moving at a constant speed, the acceleration is zero, and the net force is also zero. This means that the sum of all the forces acting on the two boxes must be zero.The force you exert on the boxes is the force of tension in the rope, and this force is equal in magnitude and opposite in direction to the force of friction on the lower box. The magnitude of the friction force on the lower box is equal to the coefficient of kinetic friction times the normal force acting on the lower box.The normal force is the force exerted by the ramp on the lower box, and it is equal in magnitude to the weight of the lower box. The weight of the lower box is equal to the mass of the lower box times the acceleration due to gravity.The magnitude of the friction force on the upper box is equal to the coefficient of static friction times the normal force acting on the upper box. The normal force acting on the upper box is equal in magnitude to the weight of the upper box.Now that we have all the forces, we can use Newton's second law to solve for the force you need to exert. The equation is:F_tension - F_friction_lower = 0

F_tension = F_friction_lowerF_friction_lower = u_k * N_lower

F_tension = u_k * m_lower * gWhere:F_tension is the force you need to exertF_friction_lower is the force of friction on the lower boxu_k is the coefficient of kinetic friction between the ramp and the lower boxN_lower is the normal force acting on the lower boxm_lower is the mass of the lower boxg is the acceleration due to gravitySubstituting the given values, we get:F_tension = (0.444) * (m_lower) * (9.8 m/s^2)The force you need to exert is therefore:F_tension = (0.444) * (m_lower) * (9.8 m/s^2)The magnitude of the friction force on the upper box is:F_friction_upper = u_s * N_upperWhere:F_friction_upper is the force of friction on the upper boxu_s is the coefficient of static friction between the two boxesN_upper is the normal force acting on the upper boxSubstituting the given values, we get:F_friction_upper = (0.800) * (m_upper) * (9.8 m/s^2)The direction of the friction force on the upper box is opposite to the direction of motion of the upper box.