Block on a Rough Incline: Is the Spring Balance Faulty?
A 4.0 kg block rests on a plane inclined at 30° to the horizontal. The coefficients of friction between block and plane are μs = 0.50 and μk = 0.40. A string attached to the block pulls parallel to the incline, upward, with a tension the student sets; take g = 10 m/s² and treat the block as a point mass. In a laboratory session the student sets the tension to 18 N, observes that the block does not move, and writes: 'The applied force is larger than the friction, so the block must be on the verge of sliding and the spring balance must be misreading.' A second student disagrees, saying the apparatus is fine and the analysis is wrong. The demonstrator asks the pair to settle the question numerically, then to predict the behaviour at 30 N and at 45 N, and finally to state the whole range of tensions for which the block stays at rest.
1. Decide whether a tension of 18 N is consistent with the block remaining at rest, and state the magnitude and direction of the friction force actually acting on the block.
Read the worked answer
Take the upward direction along the incline as positive.
- Weight component along the plane: $mg\sin30^\circ = 4.0\times10\times0.5 = 20$ N, directed down the plane.
- Normal reaction: $N = mg\cos30^\circ = 40\times0.866 = 34.6$ N.
- Limiting static friction: $f_{s,\max} = \mu_s N = 0.50\times34.6 = 17.3$ N.
- Kinetic friction (only relevant if sliding occurs): $f_k = 0.40\times34.6 = 13.9$ N.
With $T = 18$ N the net driving force, friction excluded, is $18-20 = -2$ N, i.e. 2 N down the plane. The block therefore tends to slide down, so static friction acts up the plane. Only 2 N is required, and 2 N < 17.3 N, so friction easily holds the block.
Result: the block is at rest with a friction force of 2 N directed up the incline, not at its limiting value.
The student's error is treating friction as always equal to $\mu N$. The quantity $\mu_s N$ is an upper bound; the actual static friction takes whatever value (up to that bound) is needed to prevent relative motion. The apparatus is behaving correctly.
2. Predict what happens at tensions of 30 N and 45 N, giving the acceleration in the case where motion occurs.
Read the worked answer
At 30 N: net driving force up the plane, friction excluded, is $30-20 = 10$ N. Static friction must therefore supply 10 N down the plane. Since $10 < 17.3$ N, the block still does not move; friction is 10 N down the incline. The expectation of sliding is wrong.
At 45 N: the driving force up the plane is $45-20 = 25$ N, which exceeds the 17.3 N limit, so static friction fails and the block slides up the incline. Friction now takes its kinetic value, 13.9 N, down the plane.
$$ma = 45 - 20 - 13.9 = 11.1 \text{ N} \Rightarrow a = \frac{11.1}{4.0} = 2.8\ \text{m s}^{-2}\ \text{up the incline.}$$
Note the asymmetry between starting and continuing motion: motion begins only above about 37.3 N, but once sliding, a force slightly above 33.9 N ($20 + 13.9$) keeps it moving, because $\mu_s > \mu_k$.
3. State the full range of tension for which the block stays at rest, and explain why this range is not centred on the 20 N weight component.
Read the worked answer
Rest requires the required friction not to exceed the limit, i.e. $|20 - T| \le 17.3$ N.
- Impending sliding down: $20 - T = 17.3 \Rightarrow T = 2.7$ N.
- Impending sliding up: $T - 20 = 17.3 \Rightarrow T = 37.3$ N.
So the block is in equilibrium for
$$2.7\ \text{N} \le T \le 37.3\ \text{N}.$$
Below 2.7 N the block slides down; above 37.3 N it slides up.
The range is symmetric about 20 N in the sense that the maximum friction is the same in both directions, but the block is not sitting at the midpoint of a symmetric situation: it is the applied tension that is bounded, and the two bounds are $20 \pm 17.3$ N. The lower bound is physically important and often missed - with a small enough upward pull the block breaks away and accelerates down the plane, because friction can no longer fully support the 20 N weight component.