16.7 Friction and the F <= mu R model — Practise with help

Work each one first, closed book. Find the normal reaction before you find any limit. Compare the driving force with $\mu R$ , then decide whether the body stays at rest or slides. Take $g = 9.8$ m s $^{-2}$ .

Warm-up: remember

Work out $0.4 \times 419.4$ .
Find the component of a weight of $441$ N acting down a slope of $18^\circ$ .
Which is larger, $136.3$ or $167.8$ ?

Answers

$0.4 \times 419.4 = 167.76$ .
$441\sin 18^\circ \approx 136.3$ N.
$167.8$ is larger.

If question 2 was hard, revise resolving a weight on a slope in Section 16.5 before you continue.

Practise with help

1. A block of mass $8$ kg rests on a rough horizontal floor. The coefficient of friction is $0.5$ . A horizontal push of $30$ N is applied. Find the friction force.

Hint

Find the normal reaction and the limit $\mu R$ first.

Next step

$R = 8 \times 9.8 = 78.4$ N, so $\mu R = 0.5 \times 78.4 = 39.2$ N. Compare the push with the limit.

Full working

The push of $30$ N is below the limit of $39.2$ N, so the block stays at rest. Static friction matches the push:

F = 30 \text{ N}

backwards.

2. The same block is pushed with $50$ N instead. Find the friction force and the acceleration.

Hint

The limit is unchanged at $39.2$ N. Compare the new push with it.

Next step

Since $50 > 39.2$ , the block slides and friction takes its greatest value. Subtract the friction from the push, then divide by the mass.

Full working

F = \mu R = 39.2 \text{ N},

a = \frac{50 - 39.2}{8} = \frac{10.8}{8} = 1.35 \text{ m s}^{-2}.

3. A block of mass $10$ kg is released on a rough slope of $20^\circ$ . The coefficient of friction is $0.6$ . Show that the block stays at rest, and find the friction force.

Hint

On a slope the normal reaction is $mg\cos\theta$ , not $mg$ .

Next step

The weight is $98$ N. The downhill component is $98\sin 20^\circ = 33.52$ N and the limit is $0.6 \times 98\cos 20^\circ = 55.25$ N.

Full working

The downhill component, $33.52$ N, is below the limit of $55.25$ N, so the block stays at rest. Friction matches the downhill component:

F = 33.52 \text{ N}

up the slope.

4. A block of mass $6$ kg is released on a rough slope of $35^\circ$ . The coefficient of friction is $0.4$ . Show that the block slides, and find its acceleration.

Hint

Compare the downhill component with the limit. If the component is larger, the block slides.

Next step

The weight is $58.8$ N. Downhill: $58.8\sin 35^\circ = 33.73$ N. Limit: $0.4 \times 58.8\cos 35^\circ = 19.27$ N. The block slides, so friction is $19.27$ N.

Full working

a = \frac{33.73 - 19.27}{6} = \frac{14.46}{6} \approx 2.41 \text{ m s}^{-2}

down the slope.

5. A block on a rough slope of $25^\circ$ is in limiting equilibrium. Find the coefficient of friction.

Hint

At limiting equilibrium the downhill component equals the greatest friction.

Next step

Write $mg\sin 25^\circ = \mu\,mg\cos 25^\circ$ , then divide.

Full working

\mu = \tan 25^\circ \approx 0.466.

The mass cancels, so the answer does not depend on the mass.

6. A block of mass $15$ kg rests on a rough horizontal floor. The coefficient of friction is $0.35$ . Find the smallest horizontal force that will start the block moving.

Hint

The block starts to move when the push reaches the limiting friction.

Next step

$R = 15 \times 9.8 = 147$ N, and the limiting friction is $0.35 \times 147$ .

Full working

F_{\min} = \mu R = 0.35 \times 147 = 51.45 \text{ N}.

7. A block of mass $10$ kg is pulled along a rough horizontal floor by a rope at $30^\circ$ to the horizontal with a tension of $60$ N. The coefficient of friction is $0.4$ . Find the normal reaction and the acceleration.

Hint

The rope has an upward component, so the normal reaction is less than the weight.

Next step

$R = 98 - 60\sin 30^\circ = 68$ N, so the limit is $0.4 \times 68 = 27.2$ N. The horizontal pull is $60\cos 30^\circ = 51.96$ N, which beats the limit.

Full working

R = 98 - 30 = 68 \text{ N}, \qquad F = 0.4 \times 68 = 27.2 \text{ N}.

a = \frac{51.96 - 27.2}{10} = \frac{24.76}{10} \approx 2.48 \text{ m s}^{-2}.

8. A block of mass $20$ kg on a rough horizontal floor is just on the point of moving when a horizontal force of $50$ N is applied. Find the coefficient of friction.

Hint

At the point of moving, the push equals the limiting friction.

Next step

$R = 20 \times 9.8 = 196$ N and $50 = \mu \times 196$ .

Full working

\mu = \frac{50}{196} \approx 0.255.