Q.196. Application of mechanics.

Question 196.
Two blocks A and B of weigh  1 kN and 2 kN respectively and are in equilibrium position as shown in the figure. If the coefficient of friction between the two blocks as well as the block B and the floor is 0.3 , find the minimum force P required to move the block B.

Answer 196.

Four forces act on body A considered as a free body

1) its weight 1000 N, downwards,
2) Tension, T in the string in the direction of the string obliquely upwards,
3) Normal reaction, R, vertically upwards, and
4) Frictional force (0.30) R to the left

Balancing vertical and horizontal forces,
Tsin30° = 1000 - R ... (1) and
Tcos30° = (0.30) R ... (2)
Taking ratio,
(1000 - R) / (0.30) R = tan30° = 1/√3
=> 1732 - 1.732R = (0.30) R
=> (2.032) R = 1732
=> R = 852 N.

Balancing forces in the vertical direction acting on block B treating it as a free body,
reaction R' on it from the floor = 2000 + 852 = 2852 N

Balancing horizontal forces on block B,
P = (0.30) (R + R') = (0.30) (852 + 2852)
=> P = 1111 N.

Link to YA!