Problem statement

Problem 1.61. Two touching bars 1 and 2 are placed on an inclined plane forming an angle α with the horizontal (Fig. 1.10). The masses of the bars are equal to m₁ and m₂, and the coefficients of friction between the inclined plane and these bars are equal to k₁ and k₂ respectively, with k₁ > k₂. Find: (a) the force of interaction of the bars in the process of motion; (b) the minimum value of the angle α at which the bars start sliding down.

Interpretation. Two touching bars 1 and 2 are placed on an inclined plane that makes an angle α with the horizontal. The masses of the bars are m₁ and m₂. The coefficients of friction between the plane and the bars are k₁ and k₂, with k₁ > k₂. Find (a) the force of interaction of the bars while they move, and (b) the smallest angle α at which the bars start to slide down.

We assume g = 9.8 m/s². The problem gives no number for it. This means all forces in newtons depend on g. It is a slider.

We assume m₁ = 3 kg, m₂ = 2 kg, k₁ = 0.5, k₂ = 0.2 and α = 30° to start. The problem uses letters only. All are sliders, so any case can be tried.

We assume bar 1 is the lower bar, as in the figure, and that static and sliding friction have the same coefficient. A real plane has a slightly larger static value, which only moves the limit angle.

Symbols: w is the acceleration down the slope, R is the push between the bars, N is the push of the plane on a bar, f is friction, and μ is the average friction coefficient (k₁m₁ + k₂m₂)/(m₁ + m₂).

The big idea in plain words

On a slope, every bar is pulled down by m g sin α and held back by friction k m g cos α. Bar 2 has less friction, so alone it would slide faster. It sits above bar 1, so it catches up and pushes it. That push is the interaction force R. If the coefficients were equal, nothing would push and R would be zero.

The easy trick: the bars move together, so treat them as one body of mass m₁ + m₂. R is an inside force, so it cancels. The pair slides when the total pull beats the total friction, and that happens when tan α passes the average coefficient μ. The mass of each bar acts as a weight on its own k.

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 sliding with the bars would need a pseudo-force and give the same answer with more writing, so it adds nothing here.

The solution step by step

Step 1, the forces. Perpendicular to the slope nothing accelerates, so N = m g cos α. Friction while sliding is f = k N = k m g cos α. Along the slope each bar has its pull m g sin α, friction f up the slope, and R.

Step 2, two equations (x down the slope, same w):

Step 3, add them. w = g sin α − g cos α (k₁m₁ + k₂m₂)/(m₁ + m₂) = × ( − × ), which is m/s² when the bars slide together.

Step 4, part (a). From the first line, R = m₁(w − g sin α + k₁g cos α). Put in w: R = m₁m₂(k₁ − k₂) g cos α / (m₁ + m₂) = N. It is positive because k₁ > k₂. It is the same for any steepness except for the cos α factor, so it shrinks as the plane gets steeper.

Step 5, part (b). The bars slide only if w > 0, which means sin α > μ cos α, so tan α > μ = . The smallest angle is α = °.

Step 6, below the limit angle. Nothing moves and w = 0. Friction is then static and only matches the pull, up to k m g cos α. Bar 2 needs R ≥ m₂g (sin α − k₂ cos α) and bar 1 needs R ≤ m₁g (k₁ cos α − sin α). So R lies between N and N when the bars are at rest. Newton's laws do not fix it more closely.

Step 7, if k₁ were below k₂. The lower bar would slide faster and the bars would separate with R = 0, each sliding alone. The page shows this in the last preset.

Five methods

1. Force method

The two equations above. It gives both w and R, so only this method finds R.

2. Energy method

When the pair slides a distance s, gravity does (m₁ + m₂) g sin α s and friction removes (k₁m₁ + k₂m₂) g cos α s. The rest is kinetic energy, ½(m₁ + m₂)v². With v² = 2 w s this gives the same w. It is Newton's law in energy form, and R does no net work on the pair, so it never appears.

3. Graph method

The left graph shows w/g against α. It stays at zero up to the limit angle and then rises, like a hockey stick. The middle graph shows R/(m₁g): a band while the bars rest, and a curve that falls with cos α once they slide. The right graph shows the parabola s = ½ w t².

4. Calculus method

Treat the pair as one body: (m₁ + m₂) d²s/dt² = (m₁ + m₂) g sin α − (k₁m₁ + k₂m₂) g cos α. The right side is constant, so s = ½ w t². Splitting the pair again gives R.

5. Check method

Units: g times a ratio is m/s², and R is mass times acceleration, so N. Limits: if k₁ = k₂, R = 0 and the bars act as one bar with friction k. If k₂ = 0, then μ = k₁m₁/(m₁ + m₂), a smaller limit than k₁, because bar 2 helps push bar 1 down. At α = 90° R = 0, because friction vanishes. Put R and w back into the bar 2 equation to verify it.

Fastest: the one-body method for w and the limit angle. Most reliable: the force method followed by the check method.

What each slider changes

Parameters

Exam traps

• Using tan α = k₁ or k₂ for the limit angle. Neither is right. The pair behaves like one bar with the mass-weighted coefficient μ, which lies between k₂ and k₁.
• Forgetting that the bars would slide at different rates alone. R exists only because they touch, and only when the lower bar has more friction.
• Using sliding friction k m g cos α for bars at rest. Static friction only matches the pull, so it can be smaller.
• Putting R on the wrong bar or with the wrong sign. R pushes bar 1 down the slope and bar 2 up the slope.
• Writing friction as k m g without cos α. The normal force on a slope is m g cos α.