Question

Two blocks A and B of weights 1 kN and 2 kN respectively are in equilibrium position as shown in Fig. 1. If the coefficient of friction between the two blocks as well as the block B and the floor is 0.3, the force (P) required to move the block B is ______​

Answer 1

Explanation:

Considering equilibrium of block B. Resolving the force along the horizontal and vertical directions: Tcos 30° –µRb = 0; T cos 30° = µRb ...(i) Rb + Tsin 30° – Wb = 0; Tsin 30° = Wb – Rb ...(ii) Dividing Equation (i) and (ii), We get tan 30° = (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 the equilibrium of block A: Resolving the forces along the horizontal and vertical directions, Fb + µRa – P = 0; P = Fb + µRa Ra – Rb – Wa = 0; Ra = Rb +Wa = 1.748 + 4 = 5.748 P = 0.437 + 0.25 × 5.748 P = 1.874N.Read more on Sarthaks.com - https://www.sarthaks.com/509430/two-blocks-and-of-weight-4kn-and-2kn-respectively-are-in-equilibrium-position-as-shown-fig