A plank of mass m1 with a bar of mass m2 placed on it lies on a smooth horizontal plane. A horizontal force growing with time t as F = at (a is constant) is applied to the bar. Find how the accelerations of the plank w1 and of the bar w2 depend on t, if the coefficient of friction between the plank and the bar is equal to k. Draw the approximate plots of these dependences.
For t up to t₀ = k g m₂(m₁ + m₂)/(a m₁), both bodies move together with w₁ = w₂ = a t/(m₁ + m₂). After t₀, the plank has a constant w₁ = k g m₂/m₁ and the bar has w₂ = (a t − k m₂g)/m₂. With the numbers on screen, t₀ = s and w₁ = m/s² after that.
Problem statement
Problem 1.65. A plank of mass m1 with a bar of mass m2 placed on it lies on a smooth horizontal plane. A horizontal force growing with time t as F = at (a is constant) is applied to the bar. Find how the accelerations of the plank w1 and of the bar w2 depend on t, if the coefficient of friction between the plank and the bar is equal to k. Draw the approximate plots of these dependences.
Interpretation. A plank of mass m₁ with a bar of mass m₂ placed on it lies on a smooth horizontal plane. A horizontal force that grows with time as F = a t (a is a constant) is applied to the bar. Find how the accelerations of the plank, w₁, and of the bar, w₂, depend on the time t, if the coefficient of friction between the plank and the bar is k. Draw approximate plots of these dependences.
We assume m₁ = 1 kg, m₂ = 1 kg, k = 0.2 and a = 1 N/s to start. The problem uses letters only. All four are sliders.
We assume g = 9.8 m/s². It sets the friction limit. It is a slider.
We assume the plank is long enough that the bar stays on it, and the static and sliding friction coefficients are both k.
Symbols: F is the force on the bar, f is the friction between the bodies, w₁ is the plank's acceleration, w₂ is the bar's acceleration, and t₀ is the moment when sliding starts.
The big idea in plain words
The bar is pushed, and the plank is only dragged along by friction. At first the push is small, and static friction is able to drag the plank with the bar, so they move as one body. Friction can only grow up to a limit, k m₂g. The push keeps growing, so a moment comes when friction cannot keep up. From then on the bar slides on the plank, and friction stays at its limit.
Because friction stops growing, the plank gets a steady push and a constant acceleration. The bar feels a push that keeps growing, so its acceleration keeps growing.
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 sitting on the plank accelerates, so it would need a pseudo-force on the bar and gives the same results with more writing. It adds nothing here.
The solution step by step
Step 1, the equations. The ground is smooth, so the only horizontal force on the plank is friction f from the bar: f = m₁w₁. On the bar, the force F = a t pulls forward and friction pulls back: F − f = m₂w₂.
Step 2, phase 1 (together). At the start the force is tiny, so the bar does not slip and w₁ = w₂ = w. Adding the two equations gives F = (m₁ + m₂)w, so w = a t/(m₁ + m₂) = t in m/s². The friction needed is f = m₁w = m₁a t/(m₁ + m₂), which grows in a straight line with t.
Step 3, the moment sliding starts. The normal force on the bar is m₂g, so static friction can be at most k m₂g = N. The needed friction reaches that limit when m₁a t₀/(m₁ + m₂) = k m₂g. So t₀ = k g m₂(m₁ + m₂)/(a m₁) = s.
Step 4, phase 2 (sliding). For t above t₀ the friction is f = k m₂g. Plank: w₁ = k m₂g/m₁ = m/s², constant. Bar: w₂ = (a t − k m₂g)/m₂ = t − in m/s².
Step 5, check the join. At t = t₀ the bar formula gives (a t₀ − k m₂g)/m₂. Using a t₀ = k g m₂(m₁ + m₂)/m₁, this is k g m₂/m₁, the same as w₁. So the plots join at t₀ without a jump, at the value w₁ = w₂ = m/s².
Step 6, the plots. w₁(t) is a straight line from zero up to t₀, then a flat line. w₂(t) is the same straight line up to t₀, then a steeper straight line (slope a/m₂ instead of a/(m₁ + m₂)). The graphs on screen use t/t₀ and w/(k g), so the shape depends only on η = m₂/m₁ = .
A separate calculation steps Newton's laws forward in 100000 small time steps up to t = 2.5 t₀, switching from static to sliding friction when the needed friction passes the limit. It gives a plank position of m and a bar position of m, against m and m from the formulas. The largest relative difference, over positions, speeds and accelerations, is .
Five methods
1. Force method
The two Newton equations above. They give both phases and the join directly, so this is the shortest route.
2. Energy method
Energy does not give the accelerations here, because friction in the sliding phase turns work into heat and the force changes with time. It can only check the end result: the work done by F equals the gain in kinetic energy plus k m₂g times the slip distance. It adds nothing to find w₁ and w₂.
3. Graph method
Draw f against t: a rising line, then flat at k m₂g. The plank's acceleration is f/m₁, so it has the same shape: rising, then flat. The bar's acceleration is (F − f)/m₂, the difference of two lines, which gives the bend at t₀.
4. Calculus method
Integrate the accelerations. The speed of the plank is the area under w₁, and its distance is the area under the speed. The page uses this to move the plank and the bar, and the sliding distance is the difference of the two. For t > t₀ the relative acceleration is w₂ − w₁ = (a/m₂)(t − t₀), so the bar slides a distance a(t − t₀)³/(6 m₂) over the plank
5. Check method
Units: a t is in newtons, so w = a t/(m₁ + m₂) is in m/s². Limits: with k = 0 the friction limit is zero and t₀ = 0, so the bar slides at once and the plank never moves. With a very large k, t₀ becomes large and the bodies move together for a long time. With m₁ very large, t₀ goes to k g m₂/a: the plank holds back the bar, and the bar slides as soon as F reaches k m₂g.
Fastest: the force method. Most reliable: the force method followed by the join check in step 5.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Growth rate a | Yes | t₀, the force at every moment, w₂ in phase 2, the drawing and readings. The scaled graphs keep their shape |
| Friction k | Yes | t₀, the friction limit, w₁ in phase 2, the drawing, the readings and the units of the graphs |
| Plank mass m₁ | Yes | t₀, both accelerations, the drawing and the shape of the graphs through η |
| Bar mass m₂ | Yes | t₀, the friction limit, both accelerations, the drawing and the shape of the graphs |
| Gravity g | Yes | t₀, the friction limit, and the accelerations in phase 2 |
| Given-answer mode | Sets a | The page finds the growth rate that makes sliding start at the chosen t₀ |
| Play, Pause, time bar | No | They run, stop or set the time from 0 to 2.5 t₀. The run holds 1.2 s when sliding starts and 1.2 s at the end, then repeats |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Growth rate of the force | a | N/s |
| Friction coefficient | k | |
| Masses of the plank and the bar | m₁, m₂ | kg, kg |
| Gravity | g | m/s² |
| Friction limit | k m₂g | N |
| Start of sliding | t₀ | s |
| Plank acceleration after t₀ | w₁ | m/s² |
Exam traps
- Giving the constant acceleration to the wrong body. After t₀, the plank has the constant acceleration k g m₂/m₁ and the bar has the growing one (a t − k m₂g)/m₂.
- Using k m₂g as the friction in the first phase. Before t₀ the friction is static and equals m₁a t/(m₁ + m₂), which is below the limit.
- Putting the friction limit as k (m₁ + m₂)g. The normal force between the bodies is the weight of the bar only, m₂g.
- Forgetting that the force acts on the bar, not on the plank. If the force acted on the plank, the answer would be different.
- Drawing a jump at t₀. Both accelerations are equal at t₀, so the plots join smoothly in value.