Sliding Blocks The two blocks are attached to each other by a massless string that is wrapped around a frictionless pulley (see below). When the bottom 4.00-kg block is pulled to the left by the constant force the top 2.00-kg To block slides across it to the right. Find the magnitude of the force necessary to move the blocks at constant speed. Assume that the coefficient of kinetic friction between all surfaces is 0.400. 19.6 N y ? 2.0 kg 12 4 P 4.0 kg uÑ, ? 12 ? Ñ2 (a) (b) (a) Each block moves at constant velocity. (b) Free-body diagrams for the blocks. Strategy We analyze the motions of the two blocks separately. The top block is subjected to a contact force exerted by the bottom block. The components of this force are the normal force N1 and the frictional force -0.400N1. Other forces on the top block are the tension Ti in the string and the weight of the top block itself, 19.6 N. The bottom block is subjected to contact forces due to the top block and due to the floor. The first contact force has components -N1 and 0.400/1, which are simply reaction forces to the contact forces that the bottom block exerts on the top block. The components of the contact force of the floor are N2 and 0.400N2. Other forces on this block are -P, the tension Ti, and the weight -39.2 N.
Solution Since the top block is moving horizontally to the right at constant velocity, its acceleration is zero in both the horizontal and the vertical directions. From Newton's second law, M = myas N1-19.6N = 0. EF = ma T-0.4001 = 0 Solving for the two unknowns, we obtain N1 = 19.6N and T = 0.40M1 = 7.84 N. The bottom block is also not accelerating, so the application of Newton's second law to this block gives >F= = m2ay EFy =m2ay T-P+0.400 N1+0.400 N2=0 The values of N1 and T were found with the first set of equations. When these values are substituted into the second set of equations, we can determine /2 and P. They are N2 =58.8 N and P=39.2N. Significance Understanding what direction in which to draw the friction force is often troublesome. Notice that each friction force labeled in the free-body diagram above acts in the direction opposite the motion of its corresponding block.