1) 4 2) 5 3) 11 + 2 √2 4) 30 Solution: (2) 5 R cos θ = 6 cos 0o + 2 √2 cos (180 - B) + 5 cos 270o R cos θ = 6 - 2 √2 cos B --- (i) R sin θ... View Article
1) 5N, 5 √2N 2) 10N, 10 √2N 3) (N / √5), (2N / √5) 4) None of these Solution: (3) (N / √5), (2N / √5) Let the 2 components of N be Nx and Ny... View Article
1) y - 2x = 0 2) 2y - x = 0 3) y + x = 0 4) y - x = 0 Solution: (2) 2y - x = 0 It is by the resolution of force. The equation of the line of... View Article
1) 5 2) 3 3) 7 4) 15 Solution: (1) 5 Let the two forces P and Q be inclined at an angle É‘. Then P + Q = 18 and resultant = 12 ---- (i) P2 +... View Article
1) cos-1 (P / Q) 2) cos-1 (- P / Q) 3) sin-1 (P / Q) 4) sin-1 (- P / Q) Solution: (2) cos-1 (- P / Q) Let É‘ be the angle between the forces... View Article
1) 30 √7 2) 30 √3 3) 20 √6 4) 25 √2 Solution: (1) 30 √7 Since the forces are in equilibrium. Thus one of force must be equal in magnitude... View Article
1) π / 4 2) π / 3 3) π / 2 4) 2π / 3 Solution: (4) 2π / 3 Let each of 2 equal forces be of magnitude P and let them be inclined at angle ɑ.... View Article
1) P + Q 2) P – Q 3) (1 / 2) √P2 + Q2 4) √[P2 + Q2] / 2 Solution: (4) √[P2 + Q2] / 2 Let the 2 forces be F1 and F2. Maximum resultant = P =... View Article