At the moment t = 0 the force F = at is applied to a small body of mass m resting on a smooth horizontal plane (a is a constant). The permanent direction of this force forms an angle α with the horizontal (Fig. 1.14). Find: (a) the velocity of the body at the moment of its breaking off the plane; (b) the distance traversed by the body up to this moment.
The body leaves the plane at t₀ = mg/(a sin α). Its speed then is v = mg² cos α/(2a sin² α) and the distance it has travelled is s = m²g³ cos α/(6a² sin³ α). With the numbers on screen, t₀ = s, v = m/s and s = m.
Problem statement
Problem 1.68. At the moment t = 0 the force F = at is applied to a small body of mass m resting on a smooth horizontal plane (a is a constant). The permanent direction of this force forms an angle α with the horizontal (Fig. 1.14). Find: (a) the velocity of the body at the moment of its breaking off the plane; (b) the distance traversed by the body up to this moment.
Interpretation. At the moment t = 0 the force F = a t is applied to a small body of mass m resting on a smooth horizontal plane (a is a constant). The permanent direction of this force forms an angle α with the horizontal. Find (a) the velocity of the body at the moment of its breaking off the plane, (b) the distance traversed by the body up to this moment.
We assume a = 20 N/s, m = 1 kg and α = 30° to start. The problem uses letters only. In the problem, a is the rate at which the force grows, in newtons per second, and the angle is α. All three are sliders.
We assume g = 9.8 m/s². It is a slider.
We assume the body stays on its base while it slides, and F is applied at the body so that it does not tip.
Symbols: F is the pulling force, N the normal force, t₀ the moment of leaving the plane, v and s the speed and distance at that moment.
The big idea in plain words
The force pulls the body forward and also a little upward. At first the upward part is small, and the plane holds the body up. As F grows with time, the upward part grows too, and the plane has to push less and less. When the upward part equals the weight, the plane pushes with nothing, and the body lifts off.
The forward part has been accelerating the body the whole time, and the plane is smooth, so nothing slows it down. Adding up this growing push over time gives the speed and then the distance.
The solution in each frame
Only the ground frame is used. The plane is fixed to the ground, so this frame is inertial and Newton's second law holds as written. No second frame is needed.
The solution step by step
Step 1, axes. Take x along the plane in the direction of the pull and y upward. The force has the parts F cos α = a t cos α along x and F sin α = a t sin α along y.
Step 2, the moment of leaving. While the body is on the plane, there is no vertical motion, so N + a t sin α − mg = 0. So N = mg − a t sin α. It decreases to zero at t₀ = mg/(a sin α) = N/( N/s × ) = s.
Step 3, the speed. Along x the plane is smooth, so m dv/dt = a t cos α. Integrating from rest gives v = a cos α t²/(2m) = t² in m/s. Putting t = t₀: v = a cos α (mg/(a sin α))²/(2m) = mg² cos α/(2a sin² α) = m/s.
Step 4, the distance. Integrating v from 0: s = a cos α t³/(6m) = t³ in metres. At t₀: s = m²g³ cos α/(6a² sin³ α) = m.
Step 5, the shape. The speed grows as t² and the distance as t³. So at half the time t₀/2, the speed is one quarter of v and the distance one eighth of s. The graphs use t/t₀, so these shapes never change. Only the units do.
A separate calculation steps the motion in 100000 small time steps. At each step it finds N = mg − a t sin α, adds the horizontal push to the speed, and stops when N reaches zero. It gives t₀ = s, v = m/s and s = m, against s, m/s and m from the formulas. The largest relative difference is .
Five methods
1. Force method
Newton's second law along the plane and across it, then integrate twice. It gives everything and is the base for the other methods.
2. Energy method
The work of F is the integral of F cos α dx, and it has to equal the kinetic energy ½ m v². Because F depends on time and not on distance, you still need x(t) first, so this method brings nothing new.
3. Graph method
Plot the horizontal force a t cos α against t. It is a straight line from zero. The area under it up to t is the momentum m v, which is a triangle: m v = ½ (a t cos α) t. This gives v at once, and it matches the integral.
4. Calculus method
Integrate m dv/dt = a t cos α for v, then v = dx/dt for x. This is steps 3 and 4.
5. Check method
Units: a t is in newtons, so a t cos α/m is in m/s². Limits: for α going to 90°, the force is vertical, cos α goes to 0, and t₀ goes to mg/a. The speed and the distance go to zero, since nothing pushes the body sideways. For α going to 0, t₀ and v become very large, because the upward part is tiny and the body takes very long to lift off.
Fastest: the graph method, with the triangle for the momentum. Most reliable: the force method followed by the numerical check.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Angle α | Yes | The force arrow and its parts, t₀, v, s and the dot on the angle graph |
| Growth rate a | Yes | t₀, the force arrow, v, s, the scale of the picture and the readings. The scaled graphs keep their shape |
| Mass m | Yes | t₀, v, s and the arrows |
| Gravity g | Yes | t₀, v, s and the weight arrow |
| Given-answer mode | Sets α | The page finds the angle that gives a chosen take-off speed, and warns if it is out of range |
| Play and time bar | No | They run and stop the time from 0 to t₀, then hold 1.2 s at take-off and repeat. The scrub bar sets the time |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Growth rate of the force | a | N/s |
| Angle of the force | α | ° |
| Mass of the body | m | kg |
| Gravity | g | m/s² |
| Time of leaving the plane | t₀ | s |
| Force at that moment | F | N |
| Speed and distance at that moment | v, s | m/s, m |
Exam traps
- Using the whole force F in the horizontal equation. Only the horizontal part, F cos α, accelerates the body along the plane.
- Taking a constant acceleration. The force grows with time, so the acceleration grows too, and v = a cos α t²/(2m) is not v = w t.
- Using the leaving condition F sin α = mg with F = constant. Here F = a t, so the condition gives a time, not a force.
- Mixing up the letter a. It is a rate in newtons per second here, not an acceleration, and the angle is α.
- Forgetting that the formulas end at t₀. After the body leaves the plane, it also moves vertically, and these results no longer apply.