Will the inclined-plane rig move? Static check before dynamics
A workshop demonstration rig has a 6.0 kg steel block on a plane inclined at 30 degrees to the horizontal, connected by a light inextensible string running parallel to the plane, over a frictionless pulley at the top edge, to a hanging mass of 8.0 kg. The coefficients of friction between block and plane are mu_s = 0.45 and mu_k = 0.35. The hanging mass starts 1.20 m above the floor and the block is 2.5 m from the pulley, so there is room to fall.
A trainee releases the system from rest, sees no immediate movement and concludes that friction holds everything because 0.45 is nearly half. The instructor is doubtful and wants the claim tested numerically before the class repeats the experiment.
Take g = 9.8 m/s2, sin 30 = 0.500, cos 30 = 0.866. Pulley and string masses are negligible and the string does not slip. The same 6.0 kg block is used in every trial.
1. Establish whether the system moves when released, showing the check that decides it.
Read the worked answer
Forces on the 6.0 kg block along the plane, taking up the plane as positive: tension T up the plane, weight component 6.0 x 9.8 x 0.500 = 29.4 N down the plane, and friction opposing the tendency of motion.
The 8.0 kg mass pulls with 8.0 x 9.8 = 78.4 N, so eliminating T gives a net driving force up the plane of 78.4 - 29.4 = 49.0 N.
Normal reaction N = 6.0 x 9.8 x 0.866 = 50.9 N. Maximum static friction = 0.45 x 50.9 = 22.9 N.
Since 49.0 N is greater than 22.9 N, friction cannot hold the block; it accelerates up the plane.
The trainee's comparison is invalid: mu_s on its own says nothing, because the deciding quantity is mu_s N measured against the net driving force. The margin here is about 2.1 times, so the conclusion survives small errors in mu_s. Motion would be prevented only if mu_s exceeded 49.0/50.9 = 0.96, which is far above steel on steel.
2. Find the acceleration of the blocks and the tension in the string once motion has started.
Read the worked answer
Once sliding, kinetic friction acts down the plane: f_k = 0.35 x 50.9 = 17.8 N.
Treating the two masses and string as one system: F_net = 78.4 - 29.4 - 17.8 = 31.2 N Total mass = 6.0 + 8.0 = 14.0 kg a = 31.2/14.0 = 2.23 m/s2, block up the plane and hanging mass down.
Tension from the hanging mass: 8.0 x 9.8 - T = 8.0 x 2.23, so T = 78.4 - 17.8 = 60.6 N.
Check with the block: T - 29.4 - 17.8 = 60.6 - 47.2 = 13.4 N, and 6.0 x 2.23 = 13.4 N. Consistent.
Sanity checks: the tension must lie between the two weights, 58.8 N and 78.4 N, and it does. The hanging mass falls 1.20 m from rest in t = sqrt(2 x 1.20/2.23) = 1.04 s, reaching 2.32 m/s - useful if the class is timing the fall with a stopwatch.
3. Find the range of hanging mass for which the rig stays at rest, and explain why 3.0 kg, not 4.0 kg, is the natural centre of that range.
Read the worked answer
With the block held in equilibrium along the plane, T = m2 g and N = 50.9 N, so up to 22.9 N of friction is available in either direction.
Impending motion up the plane, that is the smallest hanging mass the block can still hold back: m2 g = 29.4 - 22.9 = 6.5 N, giving m2 = 0.66 kg.
Impending motion down the plane, that is the largest hanging mass the block can hold: m2 g = 29.4 + 22.9 = 52.3 N, giving m2 = 5.34 kg.
The rig stays at rest for 0.66 kg <= m2 <= 5.34 kg.
The centre of this band is 3.0 kg because the weight component along the plane is 29.4 N, equivalent to a 3.0 kg hanging mass. Friction merely widens the band symmetrically by 22.9 N, or 2.34 kg, on each side. The 4.0 kg figure would be the balance point only on a frictionless 42 degree plane; it is a common error to use the block mass or a naive fraction of it.
Since 8.0 kg lies well outside the band, this rig will always accelerate, which is fine for demonstrating Newton's second law but unsuitable for a statics demonstration.