    Question

# In the diagram shown, the blocks A, B and C weight 3 kg, 4 kg and 5 kg respectively. The coefficient of sliding friction between any two surfaces is 0.25. A is held at rest by a massless rigid rod fixed to the wall while B and C are connected by a light flexible cord passing round a frictionless pulley. The force F necessary to drag C along the horizontal surface to the left at constant speed is (Assume that the arrangement shown in the diagram, B on C and A on B is maintained all through and g=10 m/s2) Open in App
Solution

## Given, mass of block A, MA=3 kg, mass of block B, MB=4 kg mass of block C, MC=5 kg μ=0.25 [ any two surfaces ] When force F is applied on C, the block C will move towards left. The free body diagram for C: As C is moving with constant speed F=f1+f2+T.....(1) Where f1 is friction between floor and block C and f2 between blocks B and C. FBD for mass B: As B moves with constant speed T=f2+f3.....(2) From equation (1) and (2), we get F=f1+f2+f2+f3 ⇒F=f1+2f2+f3 Substituting the values of f1, f2 and f3, we get ⇒F=f1+2f2+f3 ⇒F=μ(MA+MB+MC)g+μ(MA+MB)g+μ(MAg) Substituting the values, we get ⇒F=0.25×(3+4+5)10+0.25×2(3+4)10+0.25×(3)×10 ∴F=72.5 N Accepted answers : 72.5 , 72.50  Suggest Corrections  0      Explore more