Problem statement

Problem 1.67. A bar of mass m is pulled by means of a thread up an inclined plane forming an angle α with the horizontal (Fig. 1.13). The coefficient of friction is equal to k. Find the angle β which the thread must form with the inclined plane for the tension of the thread to be minimum. What is it equal to?

Interpretation. A bar of mass m is pulled by means of a thread up an inclined plane forming an angle α with the horizontal. The coefficient of friction is equal to k. Find the angle β which the thread must form with the inclined plane for the tension of the thread to be minimum. What is the minimum tension equal to?

We assume α = 30°, k = 0.25 and m = 2 kg to start. The problem uses letters only. All three are sliders.

We assume g = 9.8 m/s² and that the bar moves up at constant speed, so its acceleration is w = 0. The acceleration is a slider, and the best angle does not depend on it.

We assume the thread is straight and the bar stays on the plane while the thread is tilted.

Symbols: T is the tension, N the normal force, f the friction, and β the angle between the thread and the incline.

The big idea in plain words

When the thread is along the slope, all of T goes into pulling the bar up. But the bar presses hard on the plane, so friction is large. If the thread tilts upward, part of T lifts the bar. That lowers the normal force, so friction falls. But a tilted thread pulls less along the slope.

The least tension is at the angle where this gain and this loss balance. The angle turns out to depend only on the friction coefficient, not on the incline, the mass or the speed.

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 up the slope and y perpendicular to it. The weight has the parts mg sin α along −x and mg cos α along −y.

Step 2, the equations. Along x: T cos β − mg sin α − f = m w. Along y: T sin β + N − mg cos α = 0, so N = mg cos α − T sin β.

Step 3, put in friction. With f = kN, the first equation becomes T cos β − mg sin α − k(mg cos α − T sin β) = m w. Collecting the T terms: T(cos β + k sin β) = m(w + g(sin α + k cos α)). So T = N/(cos β + k sin β). For w = 0 this is T = mg(sin α + k cos α)/(cos β + k sin β). At the angle on screen, T = N.

Step 4, minimise. The top of the fraction does not depend on β. So T is least when cos β + k sin β is largest. Its derivative is −sin β + k cos β, which is zero when tan β = k. The derivative changes from positive to negative there, so it is a maximum. So β = arctan k = °.

Step 5, the value. At tan β = k, cos β = 1/√(1 + k²) and sin β = k/√(1 + k²). So cos β + k sin β = √(1 + k²) = . Then T_min = m(w + g(sin α + k cos α))/√(1 + k²) = / = N.

Step 6, does the bar stay on the plane? At the best angle the normal force is N = mg cos α − T sin β. It is positive for ordinary values, so the formulas hold. If β is made large, T sin β can pass mg cos α and the bar leaves the plane. The graphs shade this region.

Five methods

1. Force method

Newton's second law along and across the slope, then minimise T over β. This is the base for every other method.

2. Energy method

Move the bar a small distance d up the slope. The work of the thread, T d cos β, pays for the gain in height, mg d sin α, and the work against friction, k N d. It gives the same equation as the force method. It does not find the best angle faster.

3. Graph method

Plot T against β. It starts at the value for a thread along the slope, dips to a lowest point, and then rises quickly for steep threads. The lowest point is the answer. The right graph shows cos β + k sin β, whose highest point is at the same angle.

4. Calculus method

Set d(cos β + k sin β)/dβ = 0. This gives tan β = k. An even shorter route writes cos β + k sin β = √(1 + k²) cos(β − φ) with tan φ = k, whose largest value is √(1 + k²) at β = φ.

5. Check method

Units: every term in T is a force. Limits: with k = 0, tan β = 0, so the thread should be along the slope, and T_min = mg sin α. With a very large k, β approaches 90°. Here the thread would nearly lift the bar and the friction would almost vanish. A level plane (α = 0) gives T_min = k m g/√(1 + k²), which is the known result for pulling a block along the floor.

Fastest: the calculus method with the cosine form. Most reliable: the force method followed by the brute-force scan of the numerical check.

What each slider changes

Control	Changes the answer?	What it changes
Thread angle β	Yes	The thread and its arrow, T, N, f, the lift-off test and the dots on the graphs. Play sweeps it
Incline angle α	Yes	The drawing, T, N, f and the least tension. Not the best angle
Friction k	Yes	The best angle, T, N, f, the least tension and all curves
Mass m	Yes	T, N, f and the arrows. Not the best angle
Gravity g	Yes	T, N and f. Not the best angle
Acceleration w	Yes	T, N, f, the motion on screen and the least tension. Not the best angle
Starting speed	No	Only the motion on screen. The forces do not depend on speed
Given-answer mode	Sets β	The page finds the thread angle for a chosen tension, on the flat or the steep side. It warns when the tension is below the least tension
Play and time bar	No	Play sweeps β from 0° to 88° in 14 s and holds 1.2 s at the best angle and at the end. The bar moves up the slope in a loop with 1.2 s pauses

Parameters

Quantity	Symbol	Value used
Angle of the incline	α	°
Friction coefficient	k
Mass of the bar	m	kg
Gravity	g	m/s²
Acceleration	w	m/s²
Thread angle	β	°
Tension, least tension	T, Tmin	N, N
Normal force, friction	N, f	N, N

Exam traps

• Writing N = mg cos α. The thread lifts the bar, so N = mg cos α − T sin β.
• Setting the thread along the slope (β = 0) as the best choice. That is best only when k = 0.
• Giving the best angle as arctan k measured from the horizontal. It is measured from the incline.
• Thinking the best angle depends on α. The incline angle only changes the top of the fraction, not the angle that makes it least.
• Forgetting the lift-off limit. If T sin β passes mg cos α, the bar leaves the plane and the formulas stop.