Chapter 12: Q6P (page 345)

Question: A scaffold of mass 60 Kg and length 5.0 m is supported in a horizontal position by a vertical cable at each end. A window washer of mass 80 kg stands at a point1.5 mfrom one end. What is the tension in (a) the nearer cable and (b) the farther cable?

Short Answer

Expert verified

Answer:

a. The tension in the nearer cable is $8.4 \times 10^{2} N$ .

b. The tension in the farther cable is $5.3 \times 10^{2} N$ .

Step by step solution

Understanding the given information

The mass of the scaffold, M = 60 kg

The length of the scaffold, I = 5.0 m

The mass of the window washer, M = 80 Kg

The distance of the window washer from one end of the scaffold, I₁=1.5 m

Concept and formula used in the given question

Using the free body diagram and condition for equilibrium, you can write the equation for torque at point B. Using this, you can find the tension . You can balance the forces in upward and downward directions as the system is at equilibrium. From this, you can get the tension . The formulas used are given below.

Newton’s second law:

$F_{n e t} = m a$

At equilibrium,

$F_{n e t} = 0$

(a) Calculation for the tension in the nearer cable

The free body diagram:

For the equilibrium, torque at point B should be balanced,

$T_{1} l = m g l_{2} + M g (\frac{l}{2})$

So, the tension in nearer cable is,

$T_{1} = \frac{m g l_{2} + M g (\frac{l}{2})}{l} = \frac{80 (9.8) (3.5) + (60) (9.8) (2.5)}{5} = 8.4 \times 10^{2} N$

Therefore, the tension in the nearer cable is $8.4 \times 10^{2} N$ .

(b) Calculation for the tension in the farther cable

At equilibrium,

$F_{n e t} = 0 Upwardforces = Downwardforces T_{1} + T_{2} = m g + M g T_{1} + T_{2} = (m + M) g T_{1} + T_{2} = (60 + 80) (9.8) T_{1} + T_{2} = 1.4 \times 10^{3} N T_{2} = 1.4 \times 10^{3} - 8.4 \times 10^{2} = 5.3 \times 10^{2} N$