Question

A uniform, 8.0-m, 1150-kg beam is hinged to a wall and supported by a thin cable attached 2.0 m from the free end of the beam. The beam is supported at an angle of 30.0^oabove the horizontal. (a) Draw a free-body diagram of the beam. (b) Find the tension in the cable. (c) How hard does the beam push inward on the wall?

Short answer

1. The diagram is drawn.
2. The tension in the cable 10100 N.
3. The force with which the beam pushes inward on the wall is 9970 N.

Step-by-step solution

The given data

Given that an 82.0 kg climber, who is 1.90 m tall and has a center of gravity of 1.1 m from his feet, rappels down a vertical cliff with his body raised 35.0^o above the horizontal. He holds the rope 1.40 m from his feet, and it makes an 25.0^o angle with the cliff face.

Mass of climber m = 82kg

Formula used

Torque$\mathit{\tau }\mathbf{=}\mathit{F}\mathit{L}$

Where F is force exerted and L is distance.

Draw a free-body diagram of the beam(a)

The free-body diagram is as follows:

Find the tension in the cable(b)

Let T be the tension needed by the rope.

The moment arm for the force T is $\left(1.4m\right)\cos 10°$.

Applying equilibrium condition on torque

This gives torque$\tau =0N·m$

Therefore,

$$\begin{gathered} T\left(6m\right)\sin {40}^o-w\left(4m\right)\cos {30}^o=0 \\ T=\frac{\left(1150kg\right)\left(9.8m/s^2\right)\left(4m\right)\cos {30}^o}{\left(6m\right)\sin {40}^o} \\ T=10100N \end{gathered}$$

The tension in the rope is 10100 N

(c)Step 5: Find the force with which the beam pushes inward on the wall

Let H_h be the force with which the beam pushes inward on the wall

Apply condition for equilibrium

The net horizontal force is zero

Thus$\sum F_x=0$

This gives

$$\begin{gathered} H_h-T\cos 10°=0 \\ {H}_h=T\cos 10° \\ {H}_h=9970N \end{gathered}$$

Hence, the force with which the beam pushed inward on the wall is 9970 N