Non-uniform boards, loaded rods and a bar on a wedge
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A bar resting on rough ground at one end and on a smooth wedge at a point: the wedge’s reaction and the coefficient of friction
Applied Mathematics Sec 3 Term 1, Unit 2, Lesson 2, Part 2
GoalFind the reaction of a smooth wedge and the coefficient of friction at the ground for a bar that rests on rough ground at one end and on a smooth wedge at a point.
New words
smooth wedge · a fixed peg or wedge with no friction: its reaction is at right angles to the barlimiting friction · friction at its greatest value, F = μN: the bar is about to slip
Explanation
A uniform bar AB rests with its end A on rough horizontal ground and a point C of it against a smooth wedge. The bar makes an angle θ with the ground and is about to slip. Four forces act: the weight W at the midpoint, the wedge’s reaction R₂ at C, at right angles to the bar because the wedge is smooth, and at A the ground’s normal reaction N upwards and friction F along the ground. If the wedge is h high, then AC = h / sin θ. Moments about A remove N and F: W × (L/2) cos θ = R₂ × AC. Then resolve: horizontal F = R₂ sin θ (R₂ pushes the bar’s top away from the wedge’s side, friction must hold it), vertical N = W − R₂ cos θ. At limiting equilibrium F = μN, so μ = F / N. The coefficient does not depend on the weight unit, and in this problem it does not depend on the weight W at all. Ground friction is horizontal, the wedge force is not: do not mix them.
Worked examples
A uniform bar AB, 2.5 m long and weighing 40 N, rests with A on rough horizontal ground and a point C on a smooth wedge 1.2 m high. The bar is about to slip when it makes an angle θ with the ground, where sin θ = 3/5 (so cos θ = 4/5). Find the reaction of the wedge and the coefficient of friction at A. The weight acts 1.25 m from A. AC = 1.2 / (3/5) = 2 m. Moments about A (anticlockwise positive; the weight turns the bar clockwise, R₂ anticlockwise): R₂ × 2 − 40 × 1.25 × (4/5) = 0, so R₂ = 20 N. Horizontal: F = 20 × 3/5 = 12 N. Vertical: N = 40 − 20 × 4/5 = 24 N. Limiting friction: μ = 12 / 24 = 0.5.
The ground under the bar of the example has μ = 0.6. Is the bar about to slip? The friction needed is 12 N. The most the ground can give is μN = 0.6 × 24 = 14.4 N. Since 12 ≤ 14.4 the bar does not slip: it is in non-limiting equilibrium, and the friction really present (12 N) is less than the limit.
Class activity
Lean a ruler on the edge of a thick book (the ‘wedge’) with its lower end on a rough desk mat. Make the ruler steeper or flatter and see when the lower end starts to slide. Which is more likely to slip, a steep ruler or a flat one? Does a coin on the ruler change the angle at which it slips? Keep everything on the desk.
Practice
1
Take the same set-up as the example (bar 2.5 m, wedge 1.2 m high, sin θ = 3/5) but a bar of weight 30 N. Find R₂, N, F and μ.
2R₂ = 30 × 1.25 × 0.8 = 30, so R₂ = 15 N. F = 15 × 0.6 = 9 N, N = 30 − 15 × 0.8 = 18 N, μ = 9 / 18 = 0.5: the same μ as before, because every force is in proportion to the weight.
2
The 30 N bar of the last question rests on a lower wedge, 0.9 m high, at the same angle. Find R₂, N, F and μ.
AC = 0.9 / 0.6 = 1.5 m. 1.5R₂ = 30 × 1.25 × 0.8 = 30, so R₂ = 20 N. F = 20 × 0.6 = 12 N, N = 30 − 20 × 0.8 = 14 N, μ = 12 / 14 = 6/7 ≈ 0.86. A lower wedge needs more friction.
3
In which direction does the reaction of a smooth wedge act on the bar, and why?
At right angles to the bar, because a smooth contact can only push straight out of the surface (no friction along it).
4
For the bar of the example, the wedge’s reaction is 20 N and the bar makes an angle with sin θ = 3/5. Find its horizontal and its vertical component.
The reaction is at right angles to the bar, so horizontal 20 sin θ = 12 N and vertical 20 cos θ = 16 N.
5
Why is AC = h / sin θ?
Triangle A, C and the foot of the wedge’s top below C has a right angle, the angle θ at A, and the opposite side h, so sin θ = h / AC.
