Wire forms a loop and passes over the small pulleys

The wire forms a loop and passes over the small pulleys at A, B, C, and D. If its end is subjected to a force of P = 50 N, determine the force in the wire and the magnitude of the resultant force that the wire exerts on each of the pulleys.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

Solution:

Let us first draw a free body diagram focusing on the force P and the wire.

Writing an equation of equilibrium for the y-axis forces will give us the value of T, which is the tension in the wire.

$+\uparrow \sum \text{F}_\text{y}\,=\,0$

$T\text{cos}\,(30^0)\,+\,T\text{cos}\,(30^0)\,-\,50\,=\,0$

$2T\text{cos}\,(30^0)\,=\,50$

(solve for T)

$T\,=\,28.87$ N

Now we can draw a free body diagram of pulleys A and D.

We only drew the free body diagram showing pulley A. Due to symmetry, both free body diagrams will show the same forces.

It is important to understand what is being shown by this free body diagram. Note that a force of 28.87 N is being applied to the pulley at an angle of 30º. This force has two components, which can be represented by taking the cosine and sine values. Also note that a force of 28.87 N is being pulled away from the pulley along the y-axis. We can write equations summing the x-axis forces and y-axis forces to find the resultant force that effects the pulley.

$F_{R_{x}}\,=\,\sum \text{F}_\text{x}$

$F_{R_{x}}\,=\,28.87\text{sin}\,(30^0)$

$F_{R_{x}}\,=\,14.43$ N

$F_{R_{y}}\,=\,\sum \text{F}_\text{y}$

$F_{R_{y}}\,=\,28.87\,-\,28.87\text{cos}\,(30^0)$

$F_{R_{y}}\,=\,3.87$ N

We can now find the resultant force.

$F_R\,=\,\sqrt{14.43^2\,+\,3.87}$

$F_R\,=\,14.9$ N

Thus, a force of 14.9 N is exerted on pulleys A and D.

Let us now focus on pulleys B and C. We will draw a free body diagram as before.

Again, we only drew the free body diagram showing pulley B. Due to symmetry, both free body diagrams will show the same forces.

From this free body diagram, we see that a tension force of 28.87 N is applied along and x and y axes. There are no components to these individual forces, thus a direct resultant can be found.

$F_R\,=\,\sqrt{28.87^2\,+\,28.87^2}$

$F_R\,=\,40.8$ N

Thus, a force of 40.8 N is exerted on pulleys B and C.