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Calculations for the force to move a block:
Two blocks A and B having weight 4KN and 2KN respectively are in equilibrium position as shown in the figure given below. Coefficient of friction for both the surfaces is same as 0.25, make calculations for the force P which is required to move the block A.
Sol.: Considering equilibrium of the block B. Resolving force along with the horizontal and vertical directions:
Tcos30° -µRb = 0;
Tcos30° = µRb ...(i)
Rb + Tsin30° - Wb = 0;
Tsin30° = Wb - Rb ...(ii)
By dividing equation (i) and (ii), We get
tan30° = (Wb - Rb)/µRb
0.5773 = (2- Rb)/0.25Rb;
0.1443Rb = 2- Rb
Rb = 1.748N
Fb = µRb = 0.25 × 1.748 = 0.437N
Considering equilibrium of block A: Resolving forces along horizontal and vertical directions,
free body digram
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