Because the picture frame is at rest, the tension in the two ropes must exactly counteract the gravitational force on the picture frame. Drawing a free body diagram we can calculate the vertical components of the tension in the ropes:

Clearly the horizontal components of the tension in the two ropes cancel exactly. In addition, the vertical components are equal in magnitude. Since F = 0 , then the vertical components of the tension in the two ropes must cancel exactly with the gravitational force: 2T _y = mgâá’2T sin 45 ^o = (5)(9.8) = 49N . Thus: T = = 34.6N . The total tension on each rope is thus 34.6N .

Though this problem seems quite complex, it can be solved by simply drawing a free body diagram for each block. Since the resulting acceleration of each block must be of the same magnitude, we will get a set of two equations with two unknowns, T and a. First we draw the free body diagram:

On block 1, there are 3 forces acting: normal force, gravitational force and tension. The gravitational force, in terms of parallel and perpendicular components, and the normal force can be easily calculated:

The normal force is simply a reaction to the perpendicular component of the gravitational force. Thus F _N = F _Gâä¥ = 84.9N . F _N and F _Gâä¥ thus cancel, and the block is left with a force of 49N down the ramp, and the tension, T, up the ramp.

On block 2, there only two forces, the gravitational force and the tension. We know that F _G = 98N , and we denote the tension by T. Using Newton's Second Law to combine the forces on block 1 and block 2, we have 2 equations and 2 unknowns, a and T:

However, we know that a ₁ and a ₂ are the same, because the two blocks are bound together by the rope. Thus we can simply equate the right side of the two equations:

With a defined value for T, we can now plug into one of the two equations to solve for the acceleration of the system:

Thus a = 2.45m/s ² . Interpreting our answer physically, we see that block 1 accelerates up the incline, while block 2 falls, both with the same acceleration of 2.45m/s ² .

We know from the last problem that block 1 experiences a net force up the incline of 24.5 N. Since friction is present, however, there will be a static frictional force counteracting this motion. F _s ^max = μ _s F _n = (.5)(84.9) = 42.5N . Because this maximum value for the frictional force exceeds the net force of 24.5 N, the frictional force will counteract the motion of the blocks, and the 2 block system will not move. Thus a = 0 and neither block will move.