Problem :
A 5kg picture frame is held up by two ropes, each inclined 45^{ o } below vertical, as shown below. What is the tension in each of the ropes?
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:
Problem :
Consider a 10kg block resting on a frictionless plane inclined 30^{ o } connected by a rope through a pulley to a 10kg block hanging free, as seen in the figure below. What is the direction and magnitude of the resulting acceleration of the 2-block system?
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:
F _{G} | = (10kg)(9.8) | = 98N | |
F _{Gâä¥} | = F _{G}cos 30^{ o } | = 84.9N | |
F _{G || } | = F _{G}sin 30^{ o } | = 49N |
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:
F | = ma | ||
10a _{1} | = T - 49 | ||
10a _{2} | = 98 - T |
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} .
Problem :
Two 10kg blocks are connected by a rope and pulley system, as in the last problem. However, there is now friction between the block and the incline, given by μ _{s} = .5 and μ _{k} = .25 . Describe the resulting acceleration.
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.
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