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:

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.

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:

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:

F_{G}

= (10kg)(9.8)

= 98N

F_{Gâä¥}

= F_{G}cos 30^{o}

= 84.9N

F_{G || }

= F_{G}sin 30^{o}

= 49N

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:

F

= ma

10a_{1}

= T - 49

10a_{2}

= 98 - 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:

T - 49 = 98 - T Thus 2T = 147 and T = 73.5N

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

10a = 73.5 - 49 = 24.5

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.