InterviewSolution
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InterviewSolution
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Block A weighing 1000 N and block B weighing 500 N are connected by flexible wire. The coefficient of friction between block A and the plane is 0.5 while that for block B and the plane is 0.2. Determine what value of inclination of the plane the system will have impending motion down the plane? [Ref. Fig.] |
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Answer» Let θ be the inclination of the plane for which motion is impending. Free body diagrams of blocks A and B are as shown in Fig. 5.8(b). Considering equilibrium of block A, Σ Forces normal to plane = 0 → N1 – 1000 cos θ = 0 or N1 = 1000 cos θ ...(i) ∴ From law of friction F1 = µ1N1 = 0.5 1000 cos θ = 500 cos θ ...(ii) Σ Forces parallel to plane = 0 → F1 – T – 1000 sin θ = 0 or T = 500 cos θ – 1000 sin θ ...(iii) Consider the equilibrium of block B, Σ Forces normal to plane = 0 → N2 – 500 cos θ = 0 or N2 = 500 cos θ ...(iv) From law of friction, F2 = µ2N2 = 0.2 500 cos θ = 100 cos θ ...(v) Σ Forces parallel to plane = 0 → F2 + T – 500 sin θ = 0 Using the values of F2 and T from eqn. (v) and eqn. (iii), 100 cos θ + 500 cos θ – 1000 sin θ – 500 sin θ = 0 600 cos θ = 1500 sin θ tan θ = \(\frac{600}{1500}\) θ = 21.8° |
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