A square lamina OABC of length 10 cm is pivoted at 'O'. Forces act at lamina as shown in figure. If lamina remains stationary, then the magnitude of F is : [figure]
- 10 N
- 0 (zero)
- 10 2 N
- 20 N
Answer
(1) 10 N
Solution
The pivot force acts at O, so only rotational equilibrium about O is needed: O = 0.
Side a = 10 cm = 0.1 m. Taking anticlockwise torque as positive:
At C: 10 N (up) passes through O-C line = 0; F (left) at height a = +Fa
At B: 10 N (up) +10a; 10 N (right) at height a -10a
At A: 10 N (right) passes along OA = 0; 10 N (down) at distance a -10a
O = Fa + 10a - 10a - 10a = 0
F = 10 N
Answer: (1)