Topic: Engineering Mechanics
A load of 70 kN is lifted using a frictionless pulley system shown in the figure below. At each pulley, cables of square cross section with side of 10 mm are used. The stress in cable "L" will be:
- A. 100 MPa
- B. 200 MPa
- C. 400 MPa
- D. 700 MPa
Answer: C. 400 MPa
Explanation: From the block and tackle diagram: If pulling tension is $P$, the lowest hook (N) pulls with $P$. The pulley A has $2P$ acting upwards from B, so hook M holds $2P$. Pulley B has $4P$ acting upwards from C, so hook L holds $4P$.
Total upward force on the block = $P + 2P + 4P = 7P$.
$7P = 70$ kN $\implies P = 10$ kN.
Tension in cable L = $4P = 40$ kN = 40,000 N.
Cross-sectional area $A = 10 \times 10 = 100~mm^2$.
Stress = $Force / Area = \frac{40000}{100} = 400~N/mm^2 = 400$ MPa.