Question

Question: In Fig.12-35, horizontal scaffold 2, with uniform mass m₂ =30.0 kg and length L₂ =2.00 m, hangs from horizontal scaffold 1, with uniform mass m₁ = 50 kg . A 20 kgbox of nails lies on scaffold 2, centered at distance d = 0.500 mfrom the left end. What is the tension Tin the cable indicated?

Short answer

Answer:

The tension in the indicated cable is 457 N.

Step-by-step solution

Understanding the given information

Mass of scaffold 1, $m_1=50\text{kg}$

Mass of lower scaffold 2, $m_2=30\text{kg}$

Length, $L_2=2.00\text{m}$

Mass of nail box, $m=20.0\text{kg}$

d = 0.500 m

Length of scaffold 1, $L_1=3.00\text{m}$

Tension on right, $T_R=196\text{N}$

Tension on left,$T_R=196\text{N}$

Concept and formula used in the given question

Applying equilibrium conditions to both scaffolds, you can get the equations in terms of the unknown tensionsT_R and T_L . By plugging in the known values, you can find the magnitudes of tensions. The equations used are given below.

Static Equilibrium conditions:

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

Calculation for the tension T  in the cable indicated

FBD of scaffold .

Taking torque at left point.

.$\left(T_R\times L_2\right)-\left(mgd\right)-\left(m_{2}g\times \left(\frac{L}{2}\right)\right)=0$

Hence,

$$\begin{gathered} T_R=196\text{N} \\ \sum F_y=0 \\ {T}_L+T_R-mg-m_{2}g=0 \\ {T}_L+196-20\times 9.8-30\times 9.8=0 \\ {T}_L=294\text{N} \end{gathered}$$

FBD of scaffold :

Considering the pivot at left end of scaffold :
$$\begin{gathered} \sum \tau =0 \\ \left(T\times L_1\right)-\left(T_L\times d\right)-\left(m_{1}g\times \left(\frac{L_1}{2}\right)\right)-T_R\left(L_1-d\right)=0 \\ \left(T\times 3\right)-\left(294\times 0.5\right)-\left(50\times 9.8\times 1.5\right)-196\times 2.5=0 \\ T=457\text{N} \end{gathered}$$

Hence, the tension in the indicated cable is 457 N.