Problem statement

Problem 1.64. The inclined plane of Fig. 1.11 forms an angle α = 30° with the horizontal. The mass ratio m₂/m₁ = η = 2/3. The coefficient of friction between the body m₁ and the inclined plane is equal to k = 0.10. The masses of the pulley and the threads are negligible. Find the magnitude and the direction of acceleration of the body m₂ when the formerly stationary system of masses starts moving.

Interpretation. The inclined plane of the figure makes an angle α = 30° with the horizontal. The mass ratio is m₂/m₁ = η = 2/3. The coefficient of friction between the body m₁ and the plane is k = 0.10. The masses of the pulley and the threads are negligible. Find the magnitude and direction of the acceleration of m₂ when the system, at rest until now, starts moving.

We assume g = 9.8 m/s² and m₁ = 3 kg. They only set the forces in newtons and the speed of the motion on screen. The acceleration in units of g does not depend on them. Both are sliders.

We assume the thread stays tight, and the pulley has no friction.

Symbols: η = m₂/m₁, T is the thread tension, w is the acceleration of m₂, and f is friction on m₁.

The big idea in plain words

First decide which way the system wants to move. Compare the pull of m₂ (its weight) with gravity along the slope on m₁ plus the largest friction. Only if the pull wins by more than the friction limit does m₂ come down. After that, the thread links the two bodies, so they share one acceleration, and the net force left over after gravity and friction accelerates both masses together.

The solution in each frame

Only the ground frame is used. It is an inertial frame, so Newton's second law holds as written. A frame moving with m₁ would need a pseudo-force and give the same thresholds with more writing, so it adds nothing here.

The solution step by step

Step 1, which way does it move? The thread pulls m₁ up the slope with up to m₂g. Gravity pulls m₁ down the slope with m₁g sin α, and sliding friction is k m₁g cos α. Compare η with sin α + k cos α = + × = . Here η = , which is (η above the upper limit means m₂ comes down).

Step 2, the two equations. The thread length is fixed, so m₂ going down by a distance means m₁ goes up the slope by the same distance. Both have the same acceleration w and the tension T is the same on both sides. For m₂: m₂g − T = m₂w. For m₁ along the slope: T − m₁g sin α − k m₁g cos α = m₁w, because the normal force is m₁g cos α.

Step 3, add the two equations. T cancels: m₂g − m₁g sin α − k m₁g cos α = (m₁ + m₂) w. Divide by m₁: w = g (η − sin α − k cos α)/(1 + η).

Step 4, put in the numbers. sin 30° = 0.5 and k cos 30° = 0.10 × 0.866 = 0.0866. So the top of the fraction is 0.6667 − 0.5 − 0.0866 = 0.0801, and the bottom is 1.6667. Then w = 0.048 g, which is about 0.05 g. On screen: w = g = m/s², T = N.

Step 5, the direction. w is positive, so m₂ accelerates downward, and in vector form the acceleration of m₂ is w = 0.05 g, pointing the same way as g.

Five methods

1. Force method

The inequalities above. They come straight from comparing the forces at the first instant, so they are the shortest route.

2. Energy method

Let m₁ move a small distance d up the slope. The system gains energy if m₂g d − m₁g sin α d − k m₁g cos α d > 0. Dividing by m₁g d gives the same inequality as part (a). It is Newton's law in energy form, and it also needs the friction limit.

3. Graph method

The left graph shows w/g against η. It is zero in a flat band, then rises on the right (m₂ down) and falls below zero on the left (m₂ up). The edges of the flat band are the answers. The right graph shows friction: a sloping line inside the band, flat at ±k outside it.

4. Calculus method

Treat the two bodies as one system with a net driving force F(η) = m₂g − m₁g sin α along the thread. Friction can cancel F as long as |F| ≤ k m₁g cos α. The boundary values of F give the same two limits.

5. Check method

Units: η is a ratio of masses, and sin α, k cos α have no units. Limits: with k = 0 both edges meet at sin α, so there is no rest band and m₂ balances m₁ only at η = sin α. With α = 90°, m₁ hangs freely, and the band is at η = 1. With k large, the lower edge becomes negative.

Fastest: the force method. Most reliable: the force method followed by the check method, by putting a value of η just inside each edge.

What each slider changes

Control	Changes the answer?	What it changes
Angle α	Yes	Both thresholds, the regime, w, T, friction, the drawing and the graphs
Friction k	Yes	Both thresholds, the width of the rest band, w, friction and the graphs
Ratio η	Yes	w, T, friction and the drawing, inside the chosen case only
Mass m₁	No	The forces in newtons and the speed of the motion on screen. Not the thresholds
Gravity g	No	w, the forces in newtons and the motion on screen. Not the thresholds
Given-answer mode	Sets η	The page finds the ratio that gives the chosen acceleration
Case buttons a, b, c	Yes	Which case is simulated. The η slider then moves only inside that case
Play and Pause	No	In cases a and b, they start and stop the real-time motion, which repeats with a 1.2 s pause. In case c, Play sweeps η across the rest band, with 1.2 s holds at the two edges

Parameters

Quantity	Symbol	Value used
Angle of the plane	α	°
Friction coefficient	k
Mass ratio	η
Block mass, hanging mass	m₁, m₂	kg, kg
Gravity	g	m/s²
Upper limit	sin α + k cos α
Lower limit	sin α − k cos α
Acceleration of m₂	w	m/s²
Tension	T	N
Friction on m₁	f	N

Exam traps

• Using the limit k m₁g cos α for the rest case. At rest friction only balances the forces and is usually smaller than its limit.
• Getting the sign of friction wrong. When m₂ comes down, friction on m₁ points down the slope. When m₂ goes up, it points up the slope.
• Writing η > sin α for case (a). That would be true only for a smooth plane. Friction shifts the edge up by k cos α.
• Forgetting that case (b) can be impossible. If k ≥ tan α, then sin α − k cos α is zero or negative, and m₁ will not slide down.
• Dropping the g in m₂g − m₁g sin α and then mixing units. Each force has the form (mass) × g.