Two blocks connected by a rope of negligible mass are being dragged by a force a

Question

Two blocks connected by a rope of negligible mass are being dragged by a force at a 26 angle above horizontal (see figure below). Suppose F = 81.0 N, m1 = 14.0 kg, m2 = 20.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.164.

Please help very confused.

Two blocks connected by a rope of negligible mass are being dragged by a force at a 26angle above horizontal (see figure below). Suppose F = 81.0 N, m1 = 14.0 kg, m2 = 20.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.164. Please help very confused.

Explanation / Answer

To find the acceleration: First, find the frictional forces that will resist the "pull" force. When you subtract these from the pull, you will get the resulting force. Divide that force by the total mass of the system. Because Force = Mass * Acceleration, Force / Mass will give us acceleration. To Find the tension: We know again that F=ma. We can also look at the resulting forces on the second block. It will have gravitational and normal forces in the up and down directions, but will only have tension and frictional forces in the horizontal direction. So, the frictional force will "fight" the tension force that pulls the block. The total forces are F= T - Ff (Where Ff is the Force of friction) Therefore, if we subtract the friction (which we've solved for already using the coefficient of friction) from the tension, we will get the total horizontal forces on the block. By putting in the known Force as well as the Force of friction, we are left with the tension.

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