Question

Question: In Fig.12-32, a horizontal scaffold, of length 2.00 m and uniform mass 50 kg , is suspended from a building by two cables. The scaffold has dozens of paint cans stacked on it at various points. The total mass of the paint cans is 75 kg . The tension in the cable at the right is 722 N . How far horizontally from thatcable is the center of mass of the system of paint cans?

Short answer

Answer:

The center of mass of the system of paint cans is 0.702 m horizontally.

Step-by-step solution

Understanding the given information

L = 2.00 m

Mass of scaffold, m_s = 50 Kg

Mass of paint cans, M = 75.0 kg

Tension in right cable, T_R=722 N

Concept and formula used in the given question

Using the concept of static equilibrium, you can write the equation for torque with the given information. Solving this, you can find the center of mass of the system. The formula used is given below,

Static Equilibrium conditions:

$$\begin{gathered} \sum F_x=0 \\ \sum F_y=0 \\ \sum \tau =0 \end{gathered}$$

Calculation how far horizontally from that cable is the center of mass of the system of paint cans 

Applying condition of static equilibrium:

$\sum \tau =0$

Considering pivot at left point.

$$\begin{gathered} -m_{s}g\left(\frac{L}{2}\right)-Mgx+T_R\times L=0 \\ \left(-50\times 9.8\times 1\right)-\left(75\times 9.8\times x\right)+\left(722\times 2\right)=0 \\ x=1.2979\text{m} \end{gathered}$$

So, the distance from the right cable is:

$$\begin{gathered} d=2-x \\ =2-1.2979 \\ =0.702m \end{gathered}$$

Hence, the center of mass of the system of paint cans is horizontally.
0.702 m