Find the accelerations of rod A and wedge B in the arrangement shown in Fig. 1.20 if the ratio of the mass of the wedge to that of the rod equals η, and the friction between all contact surfaces is negligible.
The rod falls with wA = g/(1 + η cot²α), and the wedge moves sideways with wB = wA cot α = g/(tan α + η cot α). With η = and α = °, wA = m/s² and wB = m/s².
Problem statement
Problem 1.76. Find the accelerations of rod A and wedge B in the arrangement shown in Fig. 1.20 if the ratio of the mass of the wedge to that of the rod equals η, and the friction between all contact surfaces is negligible.
Interpretation. Find the accelerations of rod A and wedge B in the arrangement shown, if the ratio of the mass of the wedge to that of the rod equals η, and there is no friction between any surfaces. The rod moves in a vertical guide and rests on the sloping face of the wedge. The wedge slides on the floor.
We assume g = 9.8 m/s². It is a slider.
We assume the rod has a mass of 1 kg, an angle α = 45° and starts 0.35 m above the floor. The mass only sets the forces in newtons, and the height only sets the time to reach the floor. All three are sliders. The case buttons are η = 0.25, 1 and 4.
We assume the rod always touches the wedge and the wedge does not tip.
Symbols: wA is the downward acceleration of the rod, wB the horizontal acceleration of the wedge, N the push between them, and α the angle of the wedge at its tip.
The big idea in plain words
The rod can only go down. The wedge can only go sideways. They touch along a slope, so a drop of the rod forces the wedge to move. A steep slope needs little sideways motion for each drop, and a shallow slope needs a lot. The push N is the only link between them, and it acts at right angles to the slope.
The solution in each frame
Only the ground frame is needed. It is inertial, so the only forces are the real ones: the weights, N and the floor. A frame fixed to the wedge would accelerate and would add a pseudo-force on the rod for no gain.
The solution step by step
Step 1, the contact. The rod drops y while the wedge moves x to the right. The rod stays on the slope, so y = x tan α. Differentiating twice, wA = wB tan α.
Step 2, the rod. Take down as positive. The weight pulls down and the push N has an upward part N cos α. So mAg − N cos α = mAwA.
Step 3, the wedge. Take right as positive. The push on the wedge has a horizontal part N sin α. The floor and the weight are vertical and cancel. So N sin α = mBwB.
Step 4, solve. From step 3, N = mBwB/sin α. Put this into step 2 and use wB = wA cot α: mAg − mBwA cot²α = mAwA. Divide by mA: wA = g/(1 + η cot²α) = m/s².
Step 5, the wedge. wB = wA/tan α = g/(tan α + η cot α) = m/s². The push is N = N.
Step 6, the limits.
- Very light wedge, η tends to 0: wA tends to g. The rod falls freely, because the wedge gives no resistance.
- Very heavy wedge, η large: wA tends to 0. The wedge will not move, so the rod rests on it.
- Steep wedge, α tends to 90°: cot α tends to 0, so wA tends to g and wB tends to 0.
- Shallow wedge, α tends to 0°: cot α is large, so wA tends to 0.
The rod always falls slower than g, because N pushes it up. With the starting height y₀ = m, the rod reaches the floor after t = √(2y₀/wA) = s, and the wedge has moved m.
A separate calculation solves the three equations (rod, wedge and contact) as one linear system. It gets wA = m/s² and wB = m/s², against and from the formulas. It then drops the rod in 400000 small steps and finds the floor after s, against s from the formula. The largest relative difference, including N, is .
Five methods
1. Force method
Newton's law for the rod and for the wedge, plus the contact rule. This is steps 1 to 5.
2. Energy method
When the rod drops y, it loses mAgy. The wedge moves y/tan α. With the rod speed v and the wedge speed v cot α, the kinetic energy is ½mAv² + ½mBv²cot²α. Setting this equal to mAgy and differentiating gives wA = g/(1 + η cot²α), the same result. No N is needed.
3. Graph method
The first graph shows wA/g against η for the chosen angle. It falls from 1 towards 0. The second graph shows both accelerations against the angle.
4. Calculus method
Both accelerations are constant, so integrate twice: y = y₀ − ½wAt² and x = ½wBt².
5. Check method
Units: g is in m/s² and the bracket has no units. Limits: see step 6. At α = 45° and η = 1, wA = g/2 and wB = g/2, which a quick sketch of the two equations confirms.
Fastest: the energy method. Most reliable: the force method with the numerical check, because it also gives N.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Mass ratio η | Yes | wA, wB, N, the times and every curve |
| Angle α | Yes | The picture, wA, wB, N and the curves |
| Mass of the rod | No | Only the forces in newtons and the block sizes |
| Start height y₀ | Not w | The picture, the time to the floor and the wedge shift |
| Gravity g | Yes | Both accelerations, the times and the forces |
| Case buttons a, b, c | Set η | a: 0.25. b: 1. c: 4 |
| Given-answer mode | Sets η | The page finds η for a chosen wA and warns if it is out of range |
| Play and time bar | No | Play runs the motion, holds 1.2 s at the end, then repeats. The bar sets the time |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Mass ratio | η | |
| Wedge angle | α | ° |
| Rod mass | mA | kg |
| Wedge mass | mB = ηmA | kg |
| Gravity | g | m/s² |
| Rod acceleration | wA | m/s² ( percent of g) |
| Wedge acceleration | wB | m/s² |
| Push between them | N | N |
Exam traps
- Writing wA = wB. The two accelerations are linked by tan α, not equal.
- Mixing up tan and cot in the link. A shallow wedge makes the wedge move far for a small drop.
- Using N = mAg cos α, which is true only for a fixed slope. Here the wedge moves, so N is smaller.
- Putting the floor force and the wedge weight into the horizontal equation. They are vertical.
- Forgetting that the guide, not the slope, keeps the rod vertical. It gives a horizontal force on the rod that does not change the vertical motion.