A system consisting of two blocks connected by a cable is as shown in the Fig. The masses of the block A and block B are 7.5 kg and 25 kg respectively. Determine the magnitude of minimum force and it’s inclination with reference to the horizontal, to be applied on block B. The block has impending motion toward the right. Take the coefficient of friction at all contact surface to be 0.28.
Answers
Given:
The masses of the block A and block B are 7.5 kg and 25 kg respectively
To Find:
the magnitude of minimum force and it’s inclination with reference to the horizontal, to be applied on block B
Explanation:
Block A: The cable is subjected to tension T due to the force acting on the blocks. The friction force acts down the plane on block A, since it has impending motion up the plane Resolving forces parallel to the plane, we get
T = Wₐ sin 50° + μ Nₓ
T = (7.5 × 9.81) cos 50° + 0.28 Nₐ
Resolving forces perpendicular to the plane, we have
Nₐ = Wₐ cos50° = (7.5 × 9.81) cos 50°
= 69.6 N
Block B: be represented as x
Resolving forces parallel to the plate we get,
T + μₓ Nₓ = P cos θ
T + 0.28 Nₓ = P cos θ
Resolving forces perpendicular to the plane ,we get
Nₓ + P sinθ = Wₓ = 25 × 9.81
Substituting the value of Nₓ and T = 69.6 N we get
69.6 + 0.28( 25 × 9.81 b - P sinθ ) = P cosθ
or P cosθ + 0.28 P sinθ = 138.27
Differentiating w.r.t to θ and equating to zero, we have
-P cosθ + 0.28 P sinθ = 0
tanθ = 0.28 i.e., θ = tan⁻¹(0.28)
= 15.64°
Substituting the value of θ and equating to zero we get,
P cos 15.64° + 0.28 P sin 15.64° = 138.27
P = 133.15 N
∴ Hence the magnitude of the minimum force = 133.15