In the arrangement of Fig. the masses m0, m1, and m2 of bodies are equal, the masses of the pulley and the threads are negligible, and there is no friction in the pulley. Find the acceleration w with which the body m0 comes down, and the tension of the thread binding together the bodies m1 and m2, if the coefficient of friction between these bodies and the horizontal surface is equal to k. Consider possible cases.
If m₀ > k(m₁ + m₂), the hanging body comes down with w = g [m₀ − k(m₁ + m₂)] / (m₀ + m₁ + m₂), and the thread between the blocks pulls with T₂ = (1 + k) m₀ m₂ g / (m₀ + m₁ + m₂). For equal masses this is w = g(1 − 2k)/3 and T = (1 + k) m g / 3. Otherwise nothing moves and w = 0. With the numbers on screen the system , w = m/s² and T₂ = N.
Problem statement
Problem 1.60. In the arrangement of Fig. the masses m0, m1, and m2 of bodies are equal, the masses of the pulley and the threads are negligible, and there is no friction in the pulley. Find the acceleration w with which the body m0 comes down, and the tension of the thread binding together the bodies m1 and m2, if the coefficient of friction between these bodies and the horizontal surface is equal to k. Consider possible cases.
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 every mass is 2 kg and k = 0.2 to start. The problem uses letters only. All are sliders, so any case can be tried. The three masses can also be set apart, which shows the general formula.
We assume the coefficient of friction is the same for the static case and the sliding case. The problem gives one k. A real table has a slightly larger static value, which only moves the limit kc.
Symbols: w is the acceleration of every body, T₁ is the tension in the thread over the pulley, T₂ is the tension in the thread between the blocks, f₁ and f₂ are the friction forces on the blocks, kc is the largest friction coefficient that still allows motion.
The big idea in plain words
The hanging body pulls the whole chain. The table drags back on the two blocks with friction. If the pull wins, all three bodies speed up together, because the threads do not stretch. If friction wins, nothing moves at all.
The easy trick is to treat the three bodies as one body of mass m₀ + m₁ + m₂. The only outside forces along the path are the weight m₀g and the friction k(m₁ + m₂)g. The tensions are inside forces, so they cancel. That gives w at once. The tensions come later from a single body.
The words "consider possible cases" matter. The formula for w can turn negative, and a negative w would mean the blocks slide backward, which friction never allows. So there are two cases, and the page shows both.
The solution in each frame
Only the ground frame is used. The ground is an inertial frame, so Newton's second law holds as written. A frame that moves with the blocks would need pseudo-forces and would give the same answer with more writing, so it adds nothing here. The air and the table are at rest in the ground frame.
The solution step by step
Step 1, the forces. On m₀: weight m₀g down and tension T₁ up. On m₁: T₁ toward the pulley, T₂ and friction away from it. On m₂: T₂ toward the pulley and friction away from it. Weight and the table's push cancel on the two blocks.
Step 2, three equations. All bodies share one acceleration w:
T₁ − T₂ − k m₁g = m₁w
T₂ − k m₂g = m₂w
Step 3, add them. m₀g − k(m₁ + m₂)g = (m₀ + m₁ + m₂) w. With the numbers: w = × ( − × ) / , which is m/s² when the system moves.
Step 4, tensions. From the third line, T₂ = m₂(w + k g) = (1 + k) m₀ m₂ g / (m₀ + m₁ + m₂) = N. From the first line, T₁ = m₀(g − w) = N.
Step 5, case 1, the system moves. This needs w > 0, that is m₀ > k(m₁ + m₂), or k < kc = m₀/(m₁ + m₂). Now kc = . For equal masses, kc = 1/2. The results are w = g(1 − 2k)/3 and T₂ = (1 + k) m g / 3.
Step 6, case 2, the system stays at rest. If k is at least kc, the pull m₀g = N is not bigger than the largest friction k(m₁ + m₂)g = N. Static friction rises to match the pull, so w = 0 and T₁ = m₀g. Friction is now an unknown, not k m g. The thread between the blocks is then only limited, not fixed: f₂ = T₂ ≤ k m₂ g and T₁ − T₂ = f₁ ≤ k m₁ g. So T₂ lies between N and N. The simplest honest answer is "w = 0 and T₂ is undetermined within these limits".
A separate calculation writes the three equations as a 3 × 3 system and solves it by elimination for w, T₁ and T₂, then steps the motion in 3000 small time steps for 3 s. It gives w = m/s², a distance of m, against m from s = ½ w t². The largest relative difference is . It shows "none" in the at-rest case, because nothing moves.
Five methods
1. Force method
The three equations above. It gives w and both tensions, so it is the only method here that finds T₂.
2. Energy method
When the blocks move a distance s, the weight of m₀ does work m₀g s and friction takes away k(m₁ + m₂)g s. The result becomes kinetic energy of the whole chain: m₀g s − k(m₁ + m₂)g s = ½(m₀ + m₁ + m₂)v². With v² = 2 w s this gives the same w. It is Newton's law in energy form, and it never shows the tensions because they do no net work on the whole chain.
3. Graph method
The left graph shows w/g against k. It is a straight line that falls to zero at kc and then stays at zero. The kink at kc is the two cases meeting. The right graph shows s against t, a parabola that gets steeper as w grows.
4. Calculus method
Treat the chain as one body: (m₀ + m₁ + m₂) d²s/dt² = m₀g − k(m₁ + m₂)g. The right side is constant, so s = ½ w t² and v = w t. The middle graph's tension lines come from splitting the chain again.
5. Check method
Units: g times a ratio of masses gives m/s². Limits: with k = 0, w = g m₀/(m₀ + m₁ + m₂), a smooth table. When m₀ is much bigger than the others, w tends to g and T₂ tends to (1 + k) m₂ g. When k reaches kc, w tends to 0, so the two cases join without a jump. At k = 1/2 with equal masses, T₂ = (1 + 1/2) m g / 3 = m g / 2, the same as the at-rest limit. Putting the answers back gives T₁ − T₂ − k m₁g = m₁w.
Fastest: the one-body energy or force method for w alone. Most reliable: the force method followed by the check method.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Hanging mass m₀ | Yes | The pull, w, both tensions, kc and all graphs |
| Block mass m₁ | Yes | The friction load, w, the tensions, kc and the graphs |
| Block mass m₂ | Yes | w, T₂, the friction on this block, kc and the graphs |
| Friction coefficient k | Yes | w, the tensions, the friction forces and whether anything moves |
| Gravity g | Yes | w and all forces in newtons. Not w/g or the tension fractions |
| Given-answer mode | Sets k | The page finds the k that gives the chosen w, and holds w at its largest possible value if the target is too big |
| Time bar and Play | No | The clock, at half speed, with 4 s holds at 1 m and 2 m of motion |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Hanging mass | m₀ | kg |
| Block next to the pulley | m₁ | kg |
| Far block | m₂ | kg |
| Friction coefficient | k | |
| Gravity | g | m/s² |
| Limit of motion | kc | |
| Acceleration | w | m/s² |
| Tension at the pulley | T₁ | N |
| Tension between the blocks | T₂ | N |
| Friction on m₁ and m₂ | f₁, f₂ | N, N |
Exam traps
- Giving w = g(1 − 2k)/3 for every k. For k at or above 1/2 this is zero or negative, and the real answer is w = 0.
- Using friction k m g when the blocks do not move. Static friction only matches the pull and can be smaller than k m g.
- Saying the tension in the pulley thread is m₀g. It equals m₀g only at rest. While moving, T₁ = m₀(g − w) is smaller.
- Giving the same tension on both sides of m₁. The thread over the pulley and the thread to m₂ pull with different forces, T₁ and T₂. Their difference is what accelerates m₁ against friction.
- Forgetting that the blocks have different loads when the masses differ: the friction on each is k m₁g and k m₂g, not equal.