Problem statement

Problem 1.84. Two bars of masses m₁ and m₂ connected by a non-deformed light spring rest on a horizontal plane. The coefficient of friction between the bars and the surface is equal to k. What minimum constant force has to be applied in the horizontal direction to the bar of mass m₁ in order to shift the other bar?

We assume the force pushes bar 1 along the line of the spring, towards bar 2. It switches on at once and then stays constant. Pulling bar 1 away from bar 2 gives the same answer.

We assume the friction that holds a bar still is as strong as the friction that acts when it slides. Both equal k times the weight.

We assume m₁ = 1 kg, m₂ = 2 kg, k = 0.3, a spring constant κ = 100 N/m and g = 9.8 m/s², because the problem gives no numbers. The first four are sliders. The answer does not depend on κ.

Symbols: x is the largest compression of the spring, F the applied force, f the friction on a bar, κ the spring constant.

The big idea in plain words

Bar 1 does not just creep forward. A constant push makes it overshoot, like a mass on a spring. It squeezes the spring twice as far as the spot where the push and the spring balance. Bar 2 shifts at the moment when this largest squeeze pushes on it as hard as friction can hold it. Energy gives the squeeze, friction gives the limit, and the two together give F.

The solution in each frame

Only the floor frame is needed. The floor is at rest, and friction is described in it.

The solution step by step

Step 1, the condition for bar 2. The spring pushes bar 2 with force κx when its compression is x. Friction can hold bar 2 only up to k m₂ g. So bar 2 is about to shift when κx = k m₂ g. This gives x = cm here.

Step 2, the largest compression of bar 1. Bar 1 starts at rest and is at rest again at the largest compression. So the total work on it is zero. The force F does work Fx. The spring does −½κx². Friction does −k m₁ g x. Fx − ½κx² − k m₁ g x = 0. Divide by x: ½κx = F − k m₁ g.

Step 3, join the two. Put κx = k m₂ g from step 1 into step 2: ½ k m₂ g = F − k m₁ g. So F = k m₁ g + ½ k m₂ g = kg(m₁ + m₂/2) = N.

Step 4, the meaning of the two parts. The first part, k m₁ g = N, is what bar 1 needs to start moving. The second part, ½ k m₂ g = N, is half of the friction limit on bar 2. The factor half comes from the overshoot.

Step 5, the cases. If F is a little below F_min, the spring never reaches the limit, and bar 2 stays. If F is above F_min, bar 2 shifts. If F is above k(m₁ + m₂)g = N, both bars keep speeding up.

Five methods

1. Force method

Follow bar 1 in time. It obeys m₁x′′ = F − k m₁ g − κx, which is a mass on a spring. The spring force reaches 2(F − k m₁ g) at the largest compression. Setting this equal to k m₂ g gives the same F.

2. Energy method

The method in steps 1 to 3. It needs no time and no speed, so it is the fastest.

3. Graph method

The first graph shows the spring force divided by the limit k m₂ g. The curve rises, peaks, and falls back. Bar 2 shifts only if the peak reaches the red line at 1. The second graph shows the shift of bar 2 against F/F_min. It is zero below 1 and positive above.

4. Calculus method

∫(F − k m₁ g − κx)dx from 0 to x equals zero. This gives Fx − k m₁ g x − ½κx² = 0 again.

5. Check method

Units: kg × m/s² is newtons. Limits: if m₂ goes to zero, F goes to k m₁ g, the force that only starts bar 1. If m₁ goes to zero, F goes to ½ k m₂ g.

Fastest: the energy method. Most reliable: the energy method with the numerical check.

What each slider changes

Control	Changes the answer?	What it changes
Mass m₁	Yes	The minimum force and the motion
Mass m₂	Yes	The minimum force and the motion
Friction k	Yes	The minimum force and the motion
Spring constant κ	No	The compression and the motion, not the minimum force
F / Fmin	No	Only the motion: bar 2 stays below 1 and shifts above 1
Given-answer mode	Sets m₂	The page finds m₂ for a chosen Fmin and warns if it would be outside 0.2 to 5 kg
Play and time bar	No	Play shows the real motion in slow motion, holds 1.2 s at the end, then repeats

Parameters

Quantity	Symbol	Value used
Mass of bar 1	m₁	kg
Mass of bar 2	m₂	kg
Friction coefficient	k
Spring constant	κ	N/m
Limit friction on bar 2	k m₂ g	N
Compression at the limit	x	cm
Applied force	F	N ( × Fmin)
Minimum force	Fmin	N

Exam traps

• Treating the two bars as one body and writing F = k(m₁ + m₂)g. That force is larger than needed.
• Setting the spring force equal to F. The spring force is up to 2(F − k m₁ g), because bar 1 overshoots.
• Forgetting the friction on bar 1 itself.
• Keeping κ in the answer. It cancels.
• Using F = k m₂ g. That ignores bar 1 and the overshoot.