Problem statement

Problem 1.66. A small body A starts sliding down from the top of a wedge (Fig. 1.12) whose base is equal to l = 2.10 m. The coefficient of friction between the body and the wedge surface is k = 0.140. At what value of the angle α will the time of sliding be the least? What will it be equal to?

Interpretation. A small body A starts sliding down from the top of a wedge whose base is equal to l = 2.10 m. The coefficient of friction between the body and the wedge surface is k = 0.140. At what value of the angle α will the time of sliding be the least? What will it be equal to?

We assume g = 9.8 m/s² and a body mass m = 1 kg. The mass only sets the forces in newtons. The acceleration and the time do not depend on it. Both are sliders.

We assume the body is small enough to be treated as a point, starts from rest at the top, and the base of the wedge stays at l when the angle changes.

Symbols: α is the angle at the bottom of the incline, w is the acceleration along the incline, s is the path along the incline, N is the normal force, f is the friction.

The big idea in plain words

Changing the angle has two opposite effects. A steeper wedge pulls the body harder along the slope, which makes it faster. But the base is fixed, so a steeper wedge also has a longer slope to cover, which makes the trip longer. Friction makes the flat wedges slow or even impossible, because the body does not slide at all if tan α is not above k.

Somewhere in between, the two effects balance, and the time is the least. The maths below finds exactly where.

The solution in each frame

Only the ground frame is used. The wedge 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, forces. Along the incline, the weight gives mg sin α down the slope. Across the incline, N = mg cos α, so the friction is f = k mg cos α up the slope.

Step 2, acceleration. Newton's second law along the slope: mg sin α − k mg cos α = m w. So w = g(sin α − k cos α) = m/s². The body slides only if this is positive, which means tan α > k, or α > °.

Step 3, the path. The slope is the hypotenuse of a right triangle with base l, so s = l/cos α = m.

Step 4, the time. From rest, s = ½ w t². So t² = 2l/(g cos α (sin α − k cos α)). Using sin α cos α = ½ sin 2α, this is t² = 2l/(g D) with D = ½ sin 2α − k cos² α. Now t = s.

Step 5, the best angle. t is least when D is largest. Differentiate: dD/dα = cos 2α + k sin 2α = 0, so tan 2α = −1/k. The angle 2α is between 90° and 180°, so α = 45° + ½ arctan k = °. The second derivative is negative there, so it is a maximum of D.

Step 6, the least time. Write D as ½√(1 + k²) sin(2α − φ) − k/2, where tan φ = k. Its largest value is D_max = ½(√(1 + k²) − k) = . So t_min = √(4l/(g(√(1 + k²) − k))) = s.

Five methods

1. Force method

Newton's law along the slope, then the time from s = ½ w t². It gives w and t at any angle, so it is the base of all other methods.

2. Energy method

The work of gravity minus the work of friction is the kinetic energy: ½ m v² = mg l tan α − k mg cos α × (l/cos α). This gives v = √(2 g l (tan α − k)). The time then follows from the average speed, t = 2s/v. It agrees with the force method and is used in the numerical check, but it does not find the best angle faster.

3. Graph method

Plot t against α. It comes down from very large values near α = arctan k, reaches a lowest point, and rises again for steep wedges. The lowest point is the answer. The right graph shows D(α) and its highest point.

4. Calculus method

Set dD/dα = 0. This gives tan 2α = −1/k. The result 2α = 98° for k = 0.140 gives α = 49°.

5. Check method

Units: l/(g D) has units of s², since D has no units. Limits: with k = 0, tan 2α is infinite, so α = 45° and t_min = √(4l/g), the classic result for a smooth wedge. As k grows, arctan k grows, so the best angle rises above 45°. For k = 0.6 it is about 60°, and it would approach 90° only for a huge k

Fastest: the calculus method, using D. Most reliable: the calculus method followed by the brute-force scan of the numerical check.

What each slider changes

Control	Changes the answer?	What it changes
Angle α	Yes	The wedge shape, the path, w, t, the arrows, the readings and the dots on the graphs. Play sweeps it
Friction k	Yes	w, t, the best angle, the least time, the picture and all curves
Base l	Yes	The path, t, the least time and the drawing. Not the best angle
Mass m	No	Only the forces in newtons. Not w, t or the best angle
Gravity g	Yes	w, t and the least time. Not the best angle
Given-answer mode	Sets α	The page finds the angle for a chosen time, on the flat or the steep side. It warns when the time is below the least time
Play and time bar	No	Play sweeps α from 20° to 80° in 14 s, holds 1.2 s at the best angle and at the end. The body slides at half speed in a loop with 1.2 s pauses

Parameters

Quantity	Symbol	Value used
Angle of the wedge	α	°
Friction coefficient	k
Base of the wedge	l	m
Mass of the body	m	kg
Gravity	g	m/s²
Acceleration	w	m/s²
Normal force, friction	N, f	N, N
Best angle, least time	α, tmin	°, s

Exam traps

• Using the length of the slope as fixed. Here the base l is fixed, so the slope is l/cos α and changes with the angle.
• Setting dw/dα = 0 to get the best angle. That would maximise the acceleration, not minimise the time.
• Forgetting the sign of tan 2α = −1/k. It is negative, so 2α lies between 90° and 180°. The calculator gives about −82° for 2α, and 180° has to be added to bring it to 98°.
• Giving 45° as the answer. That is only for k = 0. Friction raises the best angle to 45° + ½ arctan k.
• Forgetting that below tan α = k the body does not slide at all, so the time is not defined.