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

Work through these on your own. Find the normal reaction first, then the limit $\mu R$ . Compare the driving force with the limit before you decide whether the body moves. Take $g = 9.8$ m s $^{-2}$ . Where a decimal is wanted, give the number of decimal places the question names.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

A block of mass $10$ kg rests on a rough horizontal floor. The coefficient of friction is $0.3$ . A horizontal push of $20$ N is applied. Find the normal reaction and the friction force.
A block of mass $10$ kg rests on a rough horizontal floor. The coefficient of friction is $0.3$ . A horizontal push of $40$ N is applied. Find the friction force and the acceleration.
A crate of mass $45$ kg is placed on a rough ramp inclined at $18^\circ$ . The coefficient of friction is $0.4$ . The crate is released from rest. Show that the crate stays at rest, and find the friction force. Take $g = 9.8$ m s $^{-2}$ .
A block on a rough slope of $30^\circ$ is in limiting equilibrium, about to slide. Find the coefficient of friction.
A block of mass $8$ kg rests on a rough horizontal floor. The coefficient of friction is $0.25$ . Find the smallest horizontal force that will start the block moving. Take $g = 9.8$ m s $^{-2}$ .
A block of mass $5$ kg is released on a rough slope of $25^\circ$ . The coefficient of friction is $0.3$ . Show that the block slides, and find its acceleration. Take $g = 9.8$ m s $^{-2}$ .
A block of mass $6$ kg is released on a rough slope of $15^\circ$ . The coefficient of friction is $0.5$ . Show that the block stays at rest, and find the friction force. Take $g = 9.8$ m s $^{-2}$ .
A block on a rough slope of $20^\circ$ is in limiting equilibrium. Find the coefficient of friction.
A block of mass $20$ kg rests on a rough horizontal floor. The coefficient of friction is $0.4$ . A horizontal force of $100$ N is applied. Find the acceleration. Take $g = 9.8$ m s $^{-2}$ .
A block of mass $15$ kg is pulled along a rough horizontal floor by a rope at $20^\circ$ to the horizontal with a tension of $80$ N. The coefficient of friction is $0.35$ . Find the acceleration, to $2$ d.p. Take $g = 9.8$ m s $^{-2}$ .
A block of mass $12$ kg on a rough horizontal floor is just on the point of moving when a horizontal force of $30$ N is applied. Find the coefficient of friction. Take $g = 9.8$ m s $^{-2}$ .
A block of mass $4$ kg slides across a rough horizontal floor. The coefficient of friction is $0.5$ . Find the deceleration of the block. Take $g = 9.8$ m s $^{-2}$ .

Answers and concise solutions

1. The normal reaction and the friction force

Answer: The vertical forces balance, so $R = 10 \times 9.8 = 98$ N. The limit on friction is $\mu R = 0.3 \times 98 = 29.4$ N. The push of $20$ N is within the limit, so the block stays at rest. Static friction matches the push exactly: $F = 20$ N.

2. The friction force and the acceleration

Answer: The limit on friction is $\mu R = 0.3 \times 98 = 29.4$ N. The push of $40$ N beats the limit, so the block slides and friction takes its greatest value, $29.4$ N. The resultant is $40 - 29.4 = 10.6$ N, so $a = \dfrac{10.6}{10} = 1.06$ m s $^{-2}$ .

3. The friction force

Answer: The weight is $441$ N. Perpendicular to the ramp, $R = 441\cos 18^\circ = 419.4$ N, so the limit on friction is $0.4 \times 419.4 = 167.8$ N. The component down the ramp is $441\sin 18^\circ = 136.3$ N. Since $136.3 \leq 167.8$ , the crate stays at rest. Friction matches the downhill component: $F = 136.3$ N up the ramp.

4. The coefficient of friction

Answer: In limiting equilibrium the downhill component equals the greatest friction: $mg\sin\theta = \mu\,mg\cos\theta$ . The mass cancels, so $\mu = \tan\theta = \tan 30^\circ \approx 0.577$ . The answer does not depend on the mass.

5. The smallest force

Answer: The normal reaction is $R = 8 \times 9.8 = 78.4$ N. The block starts to move when the push reaches the limiting friction, so the smallest force is $\mu R = 0.25 \times 78.4 = 19.6$ N.

6. The acceleration

Answer: The weight is $49$ N. The component down the slope is $49\sin 25^\circ = 20.71$ N and the normal reaction is $49\cos 25^\circ = 44.41$ N. The greatest friction is $0.3 \times 44.41 = 13.32$ N. Since $20.71 > 13.32$ , the block slides. The resultant is $20.71 - 13.32 = 7.39$ N, so $a = \dfrac{7.39}{5} \approx 1.48$ m s $^{-2}$ .

7. The friction force

Answer: The weight is $58.8$ N. The component down the slope is $58.8\sin 15^\circ = 15.22$ N and the normal reaction is $58.8\cos 15^\circ = 56.79$ N. The greatest friction is $0.5 \times 56.79 = 28.40$ N. Since $15.22 \leq 28.40$ , the block stays at rest, and friction takes the smaller value $15.22$ N up the slope.

8. The coefficient of friction

Answer: At limiting equilibrium $\mu = \tan\theta = \tan 20^\circ \approx 0.364$ . The block is about to slide down the slope.

9. The acceleration

Answer: The normal reaction is $R = 20 \times 9.8 = 196$ N, so the greatest friction is $0.4 \times 196 = 78.4$ N. The push of $100$ N beats the limit, so the block slides and friction is $78.4$ N. The resultant is $100 - 78.4 = 21.6$ N, so $a = \dfrac{21.6}{20} = 1.08$ m s $^{-2}$ .

10. The acceleration

Answer: Resolve vertically: $R = 15 \times 9.8 - 80\sin 20^\circ = 147 - 27.36 = 119.64$ N, so the greatest friction is $0.35 \times 119.64 = 41.87$ N. The horizontal pull is $80\cos 20^\circ = 75.18$ N, which beats the limit, so the block slides. The resultant is $75.18 - 41.87 = 33.31$ N, so $a = \dfrac{33.31}{15} \approx 2.22$ m s $^{-2}$ .

11. The coefficient of friction

Answer: The normal reaction is $R = 12 \times 9.8 = 117.6$ N. At the point of moving the push equals the limiting friction, so $\mu = \dfrac{30}{117.6} \approx 0.255$ .

12. The deceleration

Answer: The normal reaction is $R = mg$ , so the friction is $F = \mu R = \mu mg$ . Newton's second law gives $ma = \mu mg$ , and the mass cancels: $a = \mu g = 0.5 \times 9.8 = 4.9$ m s $^{-2}$ opposing the sliding. The deceleration does not depend on the mass.