A uniform meter scale of mass 0·24 kg is made of steel. It is kept on two wedges, W1 and W2, in a horizontal position. W1 is at a distance of 0·2 m from one of its ends, while W2 is at distance of 0·4 m from the other end. If the force on the scale is N1 due to W1 and N2 due to W2, then: (take g = 10·0 m s⁻²)
- (a)N1 = 1·6 N and N2 = 0·8 N
- (b)N1 = 0·8 N and N2 = 1·6 N
- (c)N1 = 0·6 N and N2 = 1·8 N
- (d)N1 = 1·8 N and N2 = 0·6 N
Correct — C, N1 = 0·6 N and N2 = 1·8 N. The scale's weight is W = mg = 0.24 x 10 = 2.4 N, acting downward at its centre (0.5 m mark). W1 supports it at the 0.2 m mark and W2 at the 0.6 m mark (0.4 m from the far end). Taking torques about W1: N2 x (0.6 - 0.2) = W x (0.5 - 0.2), i.e. N2 x 0.4 = 2.4 x 0.3, giving N2 = 1.8 N. Vertical balance then gives N1 = W - N2 = 2.4 - 1.8 = 0.6 N. The support nearer the centre of mass (W2) carries the larger share, so N2 > N1.
- (a)N1 = 1·6 N and N2 = 0·8 N — These two do add to 2.4 N, but they put the larger force on W1. W1 is farther from the centre of mass (0.3 m away) than W2 (0.1 m away), so W1 must carry the smaller load — the assignment is reversed.
- (b)N1 = 0·8 N and N2 = 1·6 N — This gets the direction right (N2 larger) but the arithmetic is off: the torque balance about W1 gives 1.8 N and 0.6 N, not 1.6 N and 0.8 N. It likely comes from mis-placing a wedge or the centre of mass.
- (d)N1 = 1·8 N and N2 = 0·6 N — This is the correct pair of numbers but attached to the wrong supports. The 1.8 N load belongs to W2 (nearer the centre), not to W1.
A rigid body resting on two supports is in equilibrium: the upward support forces balance the weight (sum of vertical forces = 0), and the turning effects cancel (sum of torques about any point = 0). For a uniform scale the whole weight can be taken to act at the geometric centre — its centre of mass.
The tidy way to solve two-support problems is to take torques about one support so that its unknown force drops out (its moment arm is zero). Solve for the second force, then use vertical balance for the first. A quick sanity check: the support closer to the centre of mass always bears the greater load. Here W2 sits only 0.1 m from the centre while W1 is 0.3 m away, so N2 must be the bigger force.
- Weight of the scale = mg = 0.24 kg x 10 m/s^2 = 2.4 N, acting at the 0.5 m (centre) mark.
- Convert 'W2 is 0.4 m from the other end' to a position: 1.0 - 0.4 = 0.6 m from the first end.
- Torque balance about W1: N2 x 0.4 = 2.4 x 0.3 -> N2 = 1.8 N.
- Vertical balance: N1 + N2 = 2.4 N -> N1 = 0.6 N.
- Forgetting to convert 'distance from the other end' into a position measured from a single reference end.
- Assigning the larger reaction to the support farther from the centre of mass instead of the nearer one.
Asked as a beam/scale on two supports, or a plank with a load, requiring the two reaction forces from torque balance.
No directly related past PYQ was found.
- practice — not a real PYQ
A uniform plank rests on two supports. The support that lies closer to the plank's centre of mass will bear:
- (a)a smaller share of the weight
- (b)a larger share of the weight
- (c)exactly half the weight
- (d)no load at all
Answer(b) a larger share of the weight — the nearer support always carries the greater reaction force.
- practice — not a real PYQ
The turning effect of a force about a point (its torque) is given by:
- (a)force x mass
- (b)force x time
- (c)force x perpendicular distance from the point
- (d)force / distance
Answer(c) force x perpendicular distance from the point — torque = F x d.