Equilibrium of a Particle and Common Forces in Mechanics – Equilibrium of concurrent forces: NEET MCQs with Answers

Explanation: Since M₂ > M₁, the 10 kg mass descends and the 5 kg mass rises, both with the same magnitude of acceleration a. For M₂: M₂g − T = M₂a For M₁: T − M₁g = M₁a Adding to eliminate T: (M₂ − M₁)g = (M₁ + M₂)a a = [(M₂ − M₁)/(M₁ + M₂)] g = [(10 − 5)/(5 + 10)] g = (5/15)g = g/3.

Explanation: Before the string is cut the whole system hangs in equilibrium, so the spring force balances the total weight: kx = (3m + m)g = 4mg A spring cannot change its extension instantaneously, so immediately after the cut the spring force is still 4mg. For block B (mass m): the string is gone, so the only force on it is its weight mg. a_B = mg/m = g, directed downwards. For block A (mass 3m): it experiences the spring force 4mg upwards and its own weight 3mg downwards. F_net = 4mg − 3mg = mg upwards a_A = mg/3m = g/3, directed upwards. Hence the accelerations of A and B are g/3 and g respectively.

Explanation: When the mass is lifted upwards with acceleration a: T₁ − mg = ma ⇒ T₁ = m(g + a) = 1(9.8 + 4.9) = 14.7 N When it is lowered with acceleration a: mg − T₂ = ma ⇒ T₂ = m(g − a) = 1(9.8 − 4.9) = 4.9 N Therefore T₁ : T₂ = 14.7 : 4.9 = 3 : 1. The ratio is exactly 3 : 1 because a happens to be exactly g/2, giving (3g/2) : (g/2).

Explanation: For a lift accelerating upwards: T − mg = ma T = m(g + a) = 1000(9.8 + 1) = 1000 × 10.8 = 10800 N. Note that 9800 N would be the tension if the lift were at rest or moving with constant velocity; the extra 1000 N is what produces the acceleration.

Explanation: Maximum tension the rope can bear: T_max = 25 × 10 = 250 N For the monkey climbing upwards with acceleration a: T − mg = ma a = (T − mg)/m = (250 − 20 × 10)/20 = (250 − 200)/20 = 50/20 = 2.5 m/s². If the monkey tried to climb faster than this the tension would exceed 250 N and the rope would snap.

Explanation: The reading of the scale is the normal reaction R it exerts on the man — his apparent weight. For upward acceleration: R − mg = ma R = m(g + a) = 80(10 + 5) = 80 × 15 = 1200 N. The man feels heavier than his true weight of 800 N. Had the lift accelerated downwards at 5 m/s², the reading would have been 80(10 − 5) = 400 N, and in free fall it would read zero.

Explanation: Applying Newton's second law to the lift, taking upwards as positive: T − mg = ma 28000 − (2000)(10) = 2000a 28000 − 20000 = 2000a 8000 = 2000a a = 4 m s⁻² Since the result is positive, the acceleration is directed upwards. This makes sense: the tension exceeds the weight, so there is a net upward force.

Explanation: The cable supports the lift together with the person, so the total mass is m = 940 + 60 = 1000 kg For upward acceleration, the free-body equation is T − mg = ma T = m(g + a) = 1000(10 + 1) = 11000 N.

Explanation: On a smooth incline the acceleration is the constant a = g sin θ, and the block starts from rest (u = 0). Distance covered in the nth second: sₙ = u + (a/2)(2n − 1) = (g sin θ/2)(2n − 1) Distance covered in the (n + 1)th second: sₙ₊₁ = (g sin θ/2)(2n + 1) Dividing, the common factor cancels: sₙ/sₙ₊₁ = (2n − 1)/(2n + 1). The angle of the incline never enters the ratio.

Explanation: There are 12 − 5 = 7 coins above the 5th coin. These coins are at rest, so together they press down on it with a force equal to their total weight. That force is 7mg.

Explanation: The sphere is in equilibrium under three forces, so their vector sum is zero. Resolving vertically gives T cos θ = W and horizontally gives T sin θ = R. Dividing these, R = W tan θ, not W cos θ, so statement (c) is wrong.

Explanation: Each hanging mass is in equilibrium, so the tension in each string is 3 × 10 = 30 N. The balance is pulled by 30 N at each end, and a spring balance reads the tension in it, not the sum of the two pulls. So it reads 30 N.

Explanation: The lower half of the rope hangs vertically and carries the full weight of 80 N. At the midpoint, the upper tension T must balance both forces: T sin θ = 60 N and T cos θ = 80 N. Hence tan θ = 60/80 = 3/4.

Explanation: The ball is in equilibrium under three forces, so their vector sum is zero. Resolving vertically gives T cos θ = W, and horizontally T sin θ = N, so N = W tan θ. Squaring and adding gives T² = N² + W². Since N and W are perpendicular, their magnitudes cannot simply be added to give T.

Explanation: Three vertical forces act on the object: the balance pull T upward, the 15 N force upward and the weight of 70 N downward. Equilibrium gives T + 15 = 70, so T = 55 N. The balance reads this tension.