Chapter 16: Review

The chapter problem, solved

The crate has a mass of $45$ kg and $g = 9.8$ m s $^{-2}$ , so its weight is

W = 45 \times 9.8 = 441 \text{ N}.

Step 1 (16.1). The crate on the floor. Two forces act on the crate: its weight $441$ N downwards and the normal reaction from the floor upwards. The crate is at rest, so by Newton's first law the resultant force is zero. The two forces are equal in size and opposite in direction, so the normal reaction is $441$ N.

Step 2 (16.2). The two workers push. The workers push with $120$ N and $95$ N in the same direction, and a resistance of $65$ N opposes the motion. The resultant along the floor is

120 + 95 - 65 = 150 \text{ N}.

Newton's second law gives

a = \frac{150}{45} = \frac{10}{3} \approx 3.33 \text{ m s}^{-2}.

Step 3 (16.3). Mass and weight. The crate's mass stays $45$ kg wherever it is. Its weight is a force that depends on $g$ : on the Earth it is $441$ N, and on the Moon, where $g = 1.6$ m s $^{-2}$ , it would be $45 \times 1.6 = 72$ N.

Step 4 (16.4). The rope over the pulley on the smooth ramp. A light inextensible string runs up the ramp, over a smooth pulley at the top, to a hanging mass of $20$ kg. The hanging weight is

20 \times 9.8 = 196 \text{ N}.

The component of the crate's weight down the ramp is

441\sin 18^\circ \approx 136.28 \text{ N}.

The hanging weight is larger, so the crate accelerates up the ramp. The resultant is

196 - 136.28 = 59.72 \text{ N},

and the total mass moved is $45 + 20 = 65$ kg, so

a = \frac{59.72}{65} \approx 0.92 \text{ m s}^{-2}.

The tension follows from the hanging mass, moving down:

20g - T = 20a, \qquad T = 20(9.8 - 0.92) \approx 177.6 \text{ N}.

Step 5 (16.5). The rope holding the crate on the smooth ramp. With the ramp smooth and the rope parallel to it, resolve along and perpendicular to the ramp.

Along the ramp, the tension balances the downhill component:

T = 441\sin 18^\circ \approx 136.3 \text{ N}.

Perpendicular to the ramp, the normal reaction balances the other component:

R = 441\cos 18^\circ \approx 419.4 \text{ N}.

Step 6 (16.6). The crate pulled along the floor. A rope at $25^\circ$ to the floor pulls with a tension of $200$ N against a resistance of $80$ N.

Resolve vertically. There is no vertical acceleration:

R + 200\sin 25^\circ = 441, \qquad R = 441 - 84.52 = 356.48 \text{ N}.

Resolve horizontally. The horizontal component of the tension is $200\cos 25^\circ = 181.26$ N, so the resultant is

181.26 - 80 = 101.26 \text{ N},

and

a = \frac{101.26}{45} \approx 2.25 \text{ m s}^{-2}.

Step 7 (16.7). The rough ramp. With the rope slack and the coefficient of friction $0.4$ , the crate is released on the ramp.

Perpendicular to the ramp, the normal reaction is still $R \approx 419.4$ N, so the greatest friction is

\mu R = 0.4 \times 419.4 \approx 167.76 \text{ N}.

The component down the ramp is $136.28$ N. Since

136.28 \leq 167.76,

the downhill component is within the limit, so the crate stays at rest. Friction takes the value it needs, and is $136.3$ N up the ramp.

Step 8 (16.8). The plank on the trestles. The uniform plank $AB$ has length $4.0$ m and mass $20$ kg, so its weight of $196$ N acts at its midpoint, $2$ m from $A$ . The crate of weight $441$ N acts $1.5$ m from $A$ .

Take moments about $A$ . The reaction at $B$ acts $4$ m from $A$ :

4R_B = 196 \times 2 + 441 \times 1.5 = 392 + 661.5 = 1053.5,

R_B = 263.375 \text{ N}.

The total load is $196 + 441 = 637$ N, so

R_A = 637 - 263.375 = 373.625 \text{ N}.

The trestle nearer the crate, at $A$ , carries the larger reaction of about $374$ N.

The chapter's answer. The crate stays at rest on the floor, with a normal reaction of $441$ N. Pushed by the two workers, it accelerates at about $3.33$ m s $^{-2}$ . Its mass is always $45$ kg, while its weight is $441$ N on the Earth. On a smooth ramp at $18^\circ$ a rope holds it with a tension of $136.3$ N and the ramp pushes back with $419.4$ N. A $20$ kg hanging mass over a smooth pulley accelerates the crate up the ramp at about $0.92$ m s $^{-2}$ with a tension of $177.6$ N. Pulled along the floor by a rope at $25^\circ$ , it accelerates at about $2.25$ m s $^{-2}$ , and the floor pushes up with $356.5$ N. On the rough ramp it stays at rest, because the greatest friction of about $167.8$ N beats the downhill pull of $136.3$ N. On the plank, the trestles carry about $374$ N and $263$ N.

Mixed practice — choose the method first

For each question, say which method you will use and why before you calculate. The items mix this chapter with earlier chapters, so the choice is not always obvious.

Forces of $9$ N and $12$ N act at right angles at a point. Find the size of the resultant and its direction from the $12$ N force.
A car of mass $1500$ kg has a resultant force of $3000$ N along the road. Find its acceleration.
Find the weight of a person of mass $75$ kg. Take $g = 9.8$ m s $^{-2}$ .
A car of mass $1200$ kg accelerates from rest to $30$ m s $^{-1}$ in $6$ s. Find its acceleration and the resultant force on it.
Masses of $5$ kg and $3$ kg hang over a smooth pulley on a light inextensible string and are released. Find the acceleration and the tension. Take $g = 9.8$ m s $^{-2}$ .
Resolve a force of $90$ N acting at $35^\circ$ above the horizontal into horizontal and vertical components, each to $2$ d.p.
A weight of $294$ N acts on a body on a slope of $20^\circ$ . Find the components along and perpendicular to the slope, each to $2$ d.p.
A block of mass $25$ kg rests on a rough horizontal floor. The coefficient of friction is $0.4$ . Find the smallest horizontal force that will start it moving. Take $g = 9.8$ m s $^{-2}$ .
A block of mass $8$ kg is released on a rough slope of $30^\circ$ . The coefficient of friction is $0.25$ . Show that the block slides, and find its acceleration, to $2$ d.p. Take $g = 9.8$ m s $^{-2}$ .
A uniform beam $AB$ has length $4$ m and mass $25$ kg. It rests on supports at $A$ and $B$ . A load of mass $40$ kg rests $1$ m from $A$ . Find the reaction at each support. Take $g = 9.8$ m s $^{-2}$ .
A force of $30$ N acts at the end of a rod of length $0.8$ m. The force makes an angle of $50^\circ$ with the rod. Find the moment about the other end, to $2$ d.p.
A body of mass $25$ kg is pulled along a smooth horizontal floor by a rope at $30^\circ$ to the horizontal with a tension of $120$ N. Find the normal reaction and the acceleration, each to $2$ d.p.
A block of mass $7$ kg rests on a smooth horizontal table. It is joined by a light inextensible string over a smooth pulley to a hanging mass of $3$ kg. Find the acceleration and the tension. Take $g = 9.8$ m s $^{-2}$ .
A stone is released from rest and falls $20$ m. Air resistance is negligible. Find its speed when it has fallen $20$ m, to $2$ d.p. Take $g = 9.8$ m s $^{-2}$ .
A particle is in equilibrium under three forces. Two of them are $(6, 0)$ N and $(-2, 5)$ N. Find the third force and its size, to $3$ d.p.
A crate of mass $20$ kg rests on a rough slope of $25^\circ$ . The coefficient of friction is $0.3$ . A light rope parallel to the slope holds the crate. Show that the crate would slide without the rope, and find the tension needed to hold it in equilibrium, to $2$ d.p. Take $g = 9.8$ m s $^{-2}$ .

Answers and concise solutions

1. The size and the direction

Answer: The size is $\sqrt{12^2 + 9^2} = 15$ N. The direction is $\tan^{-1}\left(\dfrac{9}{12}\right) = 36.87^\circ$ from the $12$ N force towards the $9$ N force.

2. The acceleration

Answer: Use $F = ma$ with the resultant force: $a = \dfrac{3000}{1500} = 2$ m s $^{-2}$ in the direction of the resultant.

3. The weight

Answer: $W = mg = 75 \times 9.8 = 735$ N acting downwards.

4. The acceleration and the resultant force

Answer: Use the motion first: $a = \dfrac{30 - 0}{6} = 5$ m s $^{-2}$ . Then $F = ma = 1200 \times 5 = 6000$ N. This mixes Chapter 15 kinematics with Newton's second law.

5. The acceleration and the tension

Answer: The acceleration is $\dfrac{(5 - 3) \times 9.8}{8} = 2.45$ m s $^{-2}$ . For the $3$ kg mass moving up, $T = 3(9.8 + 2.45) = 36.75$ N.

6. The two components

Answer: Horizontal component $= 90\cos 35^\circ = 73.72$ N. Vertical component $= 90\sin 35^\circ = 51.62$ N. The angle is measured from the horizontal, so the cosine takes the horizontal part.

7. The two components

Answer: Along the slope: $294\sin 20^\circ = 100.54$ N. Perpendicular to the slope: $294\cos 20^\circ = 276.28$ N. On a smooth slope the normal reaction is the perpendicular component.

8. The smallest force

Answer: The normal reaction is $R = 25 \times 9.8 = 245$ N, so the limiting friction is $0.4 \times 245 = 98$ N. The block starts to move when the push reaches $98$ N.

9. The acceleration

Answer: The weight is $78.4$ N. Down the slope: $78.4\sin 30^\circ = 39.2$ N. Normal reaction: $78.4\cos 30^\circ = 67.90$ N, so the greatest friction is $0.25 \times 67.90 = 16.97$ N. Since $39.2 > 16.97$ , the block slides. The resultant is $39.2 - 16.97 = 22.23$ N, so $a \approx 2.78$ m s $^{-2}$ .

10. The two reactions

Answer: The beam weight is $245$ N at $2$ m from $A$ and the load is $392$ N at $1$ m. Taking moments about $A$ : $4R_B = 245 \times 2 + 392 \times 1 = 882$ , so $R_B = 220.5$ N. The total is $637$ N, so $R_A = 637 - 220.5 = 416.5$ N.

11. The moment

Answer: The perpendicular distance is $0.8\sin 50^\circ = 0.613$ m. The moment is $30 \times 0.613 \approx 18.39$ N m.

12. The normal reaction and the acceleration

Answer: The upward component of the pull is $120\sin 30^\circ = 60$ N, so $R = 245 - 60 = 185$ N. The horizontal component is $120\cos 30^\circ = 103.92$ N, so $a = \dfrac{103.92}{25} \approx 4.16$ m s $^{-2}$ .

13. The acceleration and the tension

Answer: The hanging weight is $29.4$ N on a total mass of $10$ kg, so $a = 2.94$ m s $^{-2}$ . The block on the table has only the tension, so $T = 7 \times 2.94 = 20.58$ N.

14. The speed

Answer: The stone accelerates at $g = 9.8$ m s $^{-2}$ . Using $v^2 = u^2 + 2as$ with $u = 0$ : $v^2 = 2 \times 9.8 \times 20 = 392$ , so $v = \sqrt{392} \approx 19.80$ m s $^{-1}$ . The mass does not matter.

15. The third force and its size

Answer: The two known forces add to $(6 - 2, 0 + 5) = (4, 5)$ N. The third force cancels it, so it is $(-4, -5)$ N. Its size is $\sqrt{4^2 + 5^2} = \sqrt{41} \approx 6.403$ N.

16. The tension

Answer: The weight is $196$ N. Down the slope: $196\sin 25^\circ = 82.83$ N. The normal reaction is $196\cos 25^\circ = 177.64$ N, so the greatest friction is $0.3 \times 177.64 = 53.29$ N. Since $82.83 > 53.29$ , the crate would slide. With the rope holding it, the crate is about to slide down, so friction acts up the slope at its greatest value: $T = 82.83 - 53.29 = 29.54$ N.

What you can now do

list the forces on a body and use Newton's first law;
use $F = ma$ with a resultant force;
separate mass from weight and use $W = mg$ ;
describe free fall and terminal velocity, and find an apparent weight in a lift;
state Newton's third law and identify an action–reaction pair;
solve connected-particles problems for a pulley, a table and a ramp;
resolve a force into two perpendicular components;
write and solve the two equilibrium equations for a particle in a plane;
find the resultant of forces that are not at right angles;
use Newton's second law in two directions;
use the model $F \leq \mu R$ and decide whether a body slides;
take moments and use the principle of moments to find forces and distances.

Your relevance reflection

Return to the prompt in the chapter opener: a time you pushed, pulled, carried or balanced something heavy. Write the forces again. Now add three things you could not have said before this chapter: how to turn a push into an acceleration with $F = ma$ , how to split a force into components, and how to balance a load by taking moments. Finish with one sentence about which idea from this chapter you would use first if you had to move something heavy safely.