Q11-49P - University Physics with Modern Physics

Question

You open a restaurant and hope to entice customers by hanging out a sign. The uniform horizontal beam supporting the sign is 1.50 m long, has a mass of 16.0 kg, and is hinged to the wall. The sign itself is uniform with a mass of 28.0 kg and overall length of 1.20 m. The two wires supporting the sign are each 32.0 cm long, are 90.0 cm apart, and are equally spaced from the middle of the sign. The cable supporting the beam is 2.00 m long.

1. What minimum tension must your cable be able to support without having a sign come crashing down?
2. What minimum vertical force must the hinge be able to support without pulling out of the wall?

Step-by-Step Solution

Answer

1. The minimum tension the cable can support is 408.6 N
2. Vertical force of 161 N must be supplied.

Step 1: The given data

Given that the uniform horizontal beam supporting the sign is 1.50 m long, has a mass of 16.0 kg, and is hinged to the wall. The sign itself is uniform with a mass of 28.0 kg and overall length of 1.20 m. The two wires supporting the sign are each 32.0 cm long, are 90.0 cm apart, and are equally spaced from the middle of the sign. The cable supporting the beam is 2.00 m long.

Mass of beam ${\mathit{m}}_{\mathbf{1}}\mathbf{=}\mathbf{16}\mathbf{\text{ kg}}$

Mass of sign, ${\mathit{m}}_{\mathbf{2}}\mathbf{=}\mathbf{28}\mathbf{\text{ kg}}$

Since the wires are symmetrically placed on either side, therefore, their tensions are equal. So Tension is ${\mathit{T}}_{\mathbf{w}}\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{1}}\mathbf{g}}{\mathbf{2}}\mathbf{=}\mathbf{137}\mathbf{ }\mathbf{\text{N}}$

Step 2: Formula used

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

Where F is force exerted and l is distance.

(a) Step 3: Find minimum tension in the cable

Since the cable is 2 m long and the beam is 1.5 m long,

$\begin{array}{rcl}\cos \theta & =& \frac{1.5 \text{m}}{2 \text{m}}\\ \theta & =& 41.4°\end{array}$

Let tension in the cable is $T_c$.

Applying the condition for equilibrium,

$\sum \tau =0$gives

$$\begin{gathered} T_c\left(\sin 41.4°\right)\left(1.50 \text{m}\right)-w_{beam}\left(0.75 \text{m}\right)-T_W\left(1.50 \text{m}\right)-T_w\left(0.6 \text{m}\right)=0 \\ {T}_c=\frac{\left(1.6 \text{m}\right)\left(9.8 {\text{m/s}}^2\right)\left(0.75 \text{m}\right)+\left(137 \text{N}\right)\left(1.5 \text{m}+0.6 \text{m}\right)}{\left(1.5 \text{m}\right)\left(\sin 41.4°\right)} \\ =408.6 \text{N} \end{gathered}$$

The minimum tension is 408.6 N.

(b) Step 4: Find minimum vertical force

Let horizontal component of the force that the cliff face exerts on the climber’s feet is F_H and vertical component is F_V .

Applying equilibrium condition

Net horizontal and vertical force is zero.

$\sum F_y=0$

$\sum F_y=0$ implies $F_v+T_c\sin 41.4°-w_{beam}-2T_w=0$

$\begin{array}{rcl}{F}_v& =& 2T_w+w_{beam}-T_c\sin 41.4°\\ & =& 2\left(137 \text{N}\right)+\left(16 \text{kg}\right)\left(9.8 {\text{m/s}}^2\right)-\left(408.6 \text{N}\right)\left(\sin 41.4°\right)\\ & =& 161 \text{N}\end{array}$

Hence, vertical component is 161 N.