Forces in two dimensions and friction – Year 13 A level maths worksheet answers to download as PDF

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1. Answer to Question 1

   Method: draw a clear force diagram, resolve every force into two perpendicular directions, and use \(\cos\) for the component next to the angle and \(\sin\) for the component opposite it.

   1. The horizontal component is next to the \(40^\circ\) angle, so use cosine. Horizontal component \(=30\cos40^\circ=22.98\approx23.0\) N. So the horizontal component is about 23.0 N.
   2. The vertical component is opposite the angle, so use sine. Vertical component \(=30\sin40^\circ=19.28\approx19.3\) N. So the vertical component is about 19.3 N.
   3. Write the horizontal part with \(\mathbf i\) and the vertical part with \(\mathbf j\). \(\mathbf F=(23.0\mathbf i+19.3\mathbf j)\) N. Check: \(\sqrt{23.0^2+19.3^2}\approx30\) N, the original magnitude.

2. Answer to Question 2

   Method: draw a clear force diagram, resolve every force into two perpendicular directions, and use \(\cos\) for the component next to the angle and \(\sin\) for the component opposite it.

   1. Resolve the 9 N force and add the components of both forces. Horizontal: \(12+9\cos60^\circ=12+4.5=16.5\) N. Vertical: \(9\sin60^\circ=7.79\) N. The 12 N force has no vertical component.
   2. Use Pythagoras on the two total components. \(R=\sqrt{16.5^2+7.794^2}=\sqrt{333}=18.2\) N. So the resultant has magnitude about 18.2 N.
   3. Use \(\tan\theta=\dfrac{\text{vertical}}{\text{horizontal}}\). \(\tan\theta=\dfrac{7.794}{16.5}\), so \(\theta=25.3^\circ\) above the \(x\)-axis. So the resultant acts at about \(25.3^\circ\) above the \(x\)-axis.

3. Answer to Question 3

   1. Add the \(\mathbf i\) components and the \(\mathbf j\) components separately. \(\mathbf F=(3\mathbf i+8\mathbf j)\) N. So the resultant is \((3\mathbf i+8\mathbf j)\) N.
   2. Use Pythagoras for the magnitude and the inverse tangent for the direction. \(|\mathbf F|=\sqrt{9+64}=\sqrt{73}=8.54\) N. \(\tan\theta=\dfrac83\), so \(\theta=69.4^\circ\). So the resultant is about 8.54 N at \(69.4^\circ\) to \(\mathbf i\).
   3. Use Newton's second law \(F=ma\) with the magnitude of the resultant. \(a=\dfrac Fm=\dfrac{8.544}2=4.27\) m s\(^{-2}\). So the particle accelerates at about 4.27 m s\(^{-2}\) in the direction of the resultant.

4. Answer to Question 4

   1. The weight acts down; \(T_1\) acts up and to the left at \(40^\circ\) to the horizontal; \(T_2\) acts up and to the right at \(50^\circ\).
   2. Horizontal: \(T_1\cos40^\circ=T_2\cos50^\circ\). Vertical: \(T_1\sin40^\circ+T_2\sin50^\circ=80\).
   3. From the first equation, \(T_1=T_2\dfrac{\cos50^\circ}{\cos40^\circ}\). Since \(\cos50^\circ=\sin40^\circ\), \(T_1=T_2\tan40^\circ\). Substituting into the second, \(T_2(\tan40^\circ\sin40^\circ+\sin50^\circ)=80\), i.e. \(T_2(0.5393+0.7660)=80\) so \(T_2=61.3\) N and \(T_1=61.28\times0.8391=51.4\) N.

5. Answer to Question 5

   Method: friction can take any value up to the limiting value \(\mu R\). The object moves only if the applied force is greater than \(\mu R\); once it moves, friction equals \(\mu R\) and acts against the motion.

   1. The floor is horizontal and there is no other vertical force, so the normal reaction balances the weight. \(R=mg=15\times9.8=147\) N. So \(R=147\) N.
   2. The box starts to move when the push is larger than the limiting friction. Limiting friction \(=\mu R=0.4\times147=58.8\) N. A force just above 58.8 N will make the box move. So the least force is 58.8 N.
   3. 100 N is more than 58.8 N, so the box moves and friction takes its limiting value. Apply \(F=ma\). Resultant \(=100-58.8=41.2\) N. \(a=\dfrac{41.2}{15}=2.75\) m s\(^{-2}\). So the box accelerates at about 2.75 m s\(^{-2}\).

6. Answer to Question 6

   Method: on a slope, resolve parallel and perpendicular to the plane. Perpendicular: \(R=mg\cos\alpha\) (plus any other perpendicular components). Parallel: use \(F=ma\), with friction \(F=\mu R\) acting against the motion or the likely motion.

   1. With no friction, the only force along the slope is the component of the weight. Along the slope: \(mg\sin25^\circ=ma\), so \(a=9.8\sin25^\circ=4.14\) m s\(^{-2}\). The mass cancels, so the acceleration does not depend on the mass.
   2. First resolve perpendicular to find \(R\), then use \(F=ma\) along the slope with friction acting up the slope. Perpendicular to the slope: \(R=mg\cos25^\circ=6\times9.8\times0.9063=53.3\) N. Along the slope: \(mg\sin25^\circ-0.3R=6a\), i.e. \(24.85-15.99=6a\), so \(a=1.48\) m s\(^{-2}\). So friction reduces the acceleration from 4.14 to about 1.48 m s\(^{-2}\).

7. Answer to Question 7

   1. Forces: weight 245 N down, normal reaction \(R\) up, friction \(F\) opposing motion, tension \(P\) along the rope at \(30^\circ\) above the horizontal.
   2. Vertically (no acceleration): \(R+P\sin30^\circ=25\times9.8=245\), so \(R=245-0.5P\).
   3. Horizontally (constant speed): \(P\cos30^\circ=F=0.15R=0.15(245-0.5P)\). So \(0.866P+0.075P=36.75\), giving \(P=\dfrac{36.75}{0.9410}=39.1\) N.
   4. At \(60^\circ\) the vertical component is larger, so the normal reaction is smaller (less friction), but the horizontal component is smaller. The overall effect is an increase of \(P\) because \(\cos60^\circ\) is much smaller. (Check: \(P=\dfrac{36.75}{0.5+0.1299}=58.3\) N.)

8. Answer to Question 8

   Method: draw a clear force diagram, resolve every force into two perpendicular directions, and use \(\cos\) for the component next to the angle and \(\sin\) for the component opposite it.

   1. Each tension has a vertical component \(T\sin35^\circ\); the horizontal components cancel by symmetry. Resolving vertically: \(2T\sin35^\circ=60\), so \(T=\dfrac{30}{\sin35^\circ}\). This is the required result.
   2. Evaluate the formula. \(T=\dfrac{30}{0.5736}=52.3\) N. So each cable has a tension of about 52.3 N.
   3. Write \(T=\dfrac{30}{\sin\theta}\) and require \(T\le40\); a larger angle gives a smaller tension. \(2T\sin\theta=60\) with \(T\le40\) requires \(\sin\theta\ge\dfrac{30}{40}=0.75\), so \(\theta\ge48.6^\circ\). So the cables must make at least \(48.6^\circ\) with the horizontal.

9. Answer to Question 9

   Method: on a slope, resolve parallel and perpendicular to the plane. Perpendicular: \(R=mg\cos\alpha\) (plus any other perpendicular components). Parallel: use \(F=ma\), with friction \(F=\mu R\) acting against the motion or the likely motion.

   1. In equilibrium the friction balances the component of the weight down the slope. \(F=mg\sin12^\circ=900\times9.8\times0.2079=1834\) N. So the handbrake must provide about 1834 N of friction.
   2. Resolve perpendicular to the slope. \(R=mg\cos12^\circ=8820\times0.9781=8627\) N. So the normal reaction is about 8627 N.
   3. Equilibrium needs \(F\le\mu R\), so \(\mu\ge\dfrac FR\). \(\mu\ge\dfrac FR=\tan12^\circ=0.2126\). So the coefficient of friction must be at least 0.213.
   4. When the car slides, friction is limiting: \(0.1R\) up the slope. \(mg\sin12^\circ-0.1R=900a\): \(1833.8-862.7=900a\), so \(a=1.08\) m s\(^{-2}\) down the slope. So the car slides down the slope with acceleration about 1.08 m s\(^{-2}\).

10. Answer to Question 10

    1. Weight 196 N down; push \(P\) along the handle (down and forward); normal reaction \(R\) up; friction opposing motion.
    2. Vertically: \(R=196+P\sin35^\circ\).
    3. Horizontally: \(P\cos35^\circ=0.3R=0.3(196+P\sin35^\circ)\). \(P(0.8192-0.1721)=58.8\), so \(P=\dfrac{58.8}{0.6471}=90.9\) N.
    4. Pull the mower instead of pushing, or raise the handle angle, so the downward component of the force is smaller and the normal reaction (and friction) is reduced.

11. Answer to Question 11

    Method: on a slope, resolve parallel and perpendicular to the plane. Perpendicular: \(R=mg\cos\alpha\) (plus any other perpendicular components). Parallel: use \(F=ma\), with friction \(F=\mu R\) acting against the motion or the likely motion.

    1. \(R=mg\cos20^\circ\). Down the slope: \(mg\sin20^\circ-0.1mg\cos20^\circ=ma\). So \(a=9.8(0.3420-0.0940)=2.43\) m s\(^{-2}\).
    2. \(v=at=2.431\times5=12.2\) m s\(^{-1}\).
    3. \(s=\tfrac12at^2=\tfrac12\times2.431\times25=30.4\) m.
    4. Air resistance acts against the motion and increases with speed, so the acceleration would reduce and the skier would reach a smaller speed.

    Answers: \(a\approx2.43\) m s\(^{-2}\), \(v\approx12.2\) m s\(^{-1}\) after 5 s, and \(s\approx30.4\) m.

12. Answer to Question 12

    1. Treat each crate separately with \(F=ma\) in its own direction of motion; friction on the table crate is \(0.25\times5g\). Hanging crate: \(3g-T=3a\). Table crate: \(T-0.25\times5g=5a\). Both crates have the same acceleration \(a\) and the same tension \(T\).
    2. Add the two equations so that \(T\) cancels. Adding: \(3g-1.25g=8a\), so \(a=\dfrac{1.75\times9.8}8=2.14\) m s\(^{-2}\). So the system accelerates at about 2.14 m s\(^{-2}\).
    3. Substitute \(a\) into the equation of the hanging crate. \(T=3(g-a)=3(9.8-2.144)=22.97\approx23.0\) N. Check with the table crate: \(12.25+5\times2.144=22.97\) N.
    4. The tension is the same throughout the string (light) and both crates have the same acceleration (inextensible).

13. Answer to Question 13

    Method: draw a clear force diagram, resolve every force into two perpendicular directions, and use \(\cos\) for the component next to the angle and \(\sin\) for the component opposite it.

    1. Resolve each tension perpendicular to the centre line. Each rope has a sideways component \(4000\sin20^\circ\) in opposite directions, so they cancel. So there is no sideways resultant and the barge moves along the centre line.
    2. Resolve each tension along the centre line and add. Forward force \(=2\times4000\cos20^\circ=7518\) N. So the total forward force is about 7518 N.
    3. Subtract the resistance, then use \(F=ma\). Net force \(=7518-6200=1318\) N, so \(a=\dfrac{1318}{90000}=0.0146\) m s\(^{-2}\). So the barge accelerates at about 0.0146 m s\(^{-2}\).

14. Answer to Question 14

    Method: for connected vehicles, use the whole system to find the acceleration (the tension is internal and cancels), then use one vehicle alone to find the tension.

    1. Total mass 2000 kg; total resistance \(400+300=700\) N. Whole system: \(3000-700=2000a\), so \(a=1.15\) m s\(^{-2}\). So the acceleration is 1.15 m s\(^{-2}\).
    2. For the caravan alone, the tow bar pulls it forward and the resistance acts backwards. Caravan: \(T-300=800\times1.15\), so \(T=1220\) N. So the tension in the tow bar is 1220 N.
    3. At a steady speed the resultant force on each part is zero. Steady speed means no acceleration: driving force \(=700\) N and for the caravan \(T=300\) N. So the engine only needs 700 N and the tow bar tension is 300 N.

15. Answer to Question 15

    Method: on a slope, resolve parallel and perpendicular to the plane. Perpendicular: \(R=mg\cos\alpha\) (plus any other perpendicular components). Parallel: use \(F=ma\), with friction \(F=\mu R\) acting against the motion or the likely motion.

    1. The horizontal force has a component \(P\sin30^\circ\) into the plane, which adds to the weight component. Perpendicular to the plane: \(R=mg\cos30^\circ+P\sin30^\circ\) with \(mg=98\). This is the required result.
    2. If the block is about to move up, friction acts down the slope and is limiting. Up the plane: \(P\cos30^\circ=98\sin30^\circ+0.3R=49+0.3(84.87+0.5P)\). So \(0.8660P-0.15P=49+25.46\), \(0.716P=74.46\) and \(P=104\) N. So \(P\approx104\) N.
    3. If the block is about to slide down, friction acts up the slope and is limiting. Down the plane: \(P\cos30^\circ+0.3R=98\sin30^\circ\): \(0.866P+25.46+0.15P=49\), so \(1.016P=23.54\) and \(P=23.2\) N. So \(P\approx23.2\) N.
    4. Between these two limiting cases friction is strong enough to hold the block. \(23.2\le P\le104\) (3 s.f.).

16. Answer to Question 16

    Method: draw a clear force diagram, resolve every force into two perpendicular directions, and use \(\cos\) for the component next to the angle and \(\sin\) for the component opposite it.

    1. Resolve vertically and horizontally; the vertical component of the tension balances the weight. Vertically: \(T\cos25^\circ=40\). Horizontally: \(T\sin25^\circ=H\). Dividing the vertical equation by \(\cos25^\circ\) gives the required result.
    2. Evaluate \(T\), then use the horizontal equation. \(T=\dfrac{40}{0.9063}=44.1\) N. \(H=44.13\times0.4226=18.7\) N, or \(H=40\tan25^\circ=18.7\) N. So \(T\approx44.1\) N and \(H\approx18.7\) N.
    3. Dividing the two equations gives \(H=40\tan\theta\). \(H=40\tan40^\circ=33.6\) N. So a bigger angle needs a bigger horizontal force.
    4. Use \(T=\dfrac{40}{\cos\theta}\le100\). \(T\cos\theta=40\) and \(T\le100\) require \(\cos\theta\ge0.4\), so \(\theta\le66.4^\circ\). So the angle to the vertical must be at most \(66.4^\circ\).

17. Answer to Question 17

    1. The pull from the hanging mass is \(3g=29.4\) N. The component of the weight down the slope is \(4g\sin30^\circ=19.6\) N and the maximum friction is \(0.2\times4g\cos30^\circ=6.79\) N. Since \(29.4>19.6+6.79\), the block moves up.
    2. \(R=4g\cos30^\circ=33.95\). Hanging: \(3g-T=3a\). Block: \(T-4g\sin30^\circ-0.2R=4a\). Adding: \(29.4-19.6-6.79=7a\), so \(a=0.430\) m s\(^{-2}\).
    3. \(T=3(g-a)=3(9.8-0.430)=28.1\) N.
    4. After 2 s the block has speed \(0.86\) m s\(^{-1}\) up the slope. With the string broken, the block decelerates with friction and gravity acting down the slope: \(4a=-(19.6+6.79)\), \(a=-6.60\) m s\(^{-2}\). It stops after \(0.13\) s. Then \(\tan30^\circ=0.577>0.2\) means it slides back down.

18. Answer to Question 18

    Method: in equilibrium the vector sum of the forces is zero, so the \(\mathbf i\) components and the \(\mathbf j\) components each add to zero.

    1. Equilibrium: \(p+2-4=0\) gives \(p=2\); \(3+q-6=0\) gives \(q=3\).
    2. Resultant of \(\mathbf F_1\) and \(\mathbf F_2\) is \((4\mathbf i+6\mathbf j)\) N, so \(\mathbf a=\dfrac{4\mathbf i+6\mathbf j}{0.5}=(8\mathbf i+12\mathbf j)\) m s\(^{-2}\).
    3. \(|\mathbf a|=\sqrt{64+144}=\sqrt{208}=14.4\) m s\(^{-2}\), at \(\tan^{-1}\dfrac{12}{8}=56.3^\circ\) above \(\mathbf i\).
    4. \(v=at=14.42\times3=43.3\) m s\(^{-1}\).

19. Answer to Question 19

    1. At the point of sliding, limiting equilibrium gives \(F=\mu R\) with \(R=mg\cos\theta\) and \(F=mg\sin\theta\). Dividing, \(\mu=\tan\theta\).
    2. Mean \(=\dfrac{17+19+18+20+18}5=18.4^\circ\), so \(\mu\approx\tan18.4^\circ=0.333\).
    3. \(\tan15^\circ=0.268<0.333\), so friction can hold the block: it does not slide (and the mass does not matter).
    4. Assumptions: the block is a particle and the coefficient of friction is constant over the surface. Improvement: repeat more trials, or use a smooth, slow rise of the board and a larger set of different blocks.

20. Answer to Question 20

    1. \(R=mg-P\sin\theta\) and \(P\cos\theta=\mu R\). So \(P\cos\theta=\mu mg-\mu P\sin\theta\), giving \(P(\cos\theta+\mu\sin\theta)=\mu mg\).
    2. \(P\) is smallest when the denominator \(\cos\theta+\mu\sin\theta\) is largest. Differentiating: \(-\sin\theta+\mu\cos\theta=0\), so \(\tan\theta=\mu\). (This is a maximum of the denominator, since the second derivative is \(-\cos\theta-\mu\sin\theta<0\).)
    3. \(\theta=\tan^{-1}0.4=21.8^\circ\). Denominator \(=\cos21.8^\circ+0.4\sin21.8^\circ=0.9285+0.1486=1.0770\). \(P_{min}=\dfrac{0.4\times30\times9.8}{1.077}=109\) N.

    Answer: the crate is easiest to pull at about \(21.8^\circ\) above the horizontal, where the tension is about 109 N.

21. Answer to Question 21

    Method: on a slope, resolve parallel and perpendicular to the plane. Perpendicular: \(R=mg\cos\alpha\) (plus any other perpendicular components). Parallel: use \(F=ma\), with friction \(F=\mu R\) acting against the motion or the likely motion.

    1. Equilibrium needs \(F=mg\sin\alpha\) with \(R=mg\cos\alpha\) and \(F\le\mu R\). So \(mg\sin\alpha\le\mu mg\cos\alpha\), i.e. \(\tan\alpha\le\mu\). If \(\tan\alpha>\mu\) the friction cannot be large enough and the block slides.
    2. \(\alpha_{max}=\tan^{-1}0.45=24.2^\circ\).
    3. \(a=g(\sin30^\circ-0.45\cos30^\circ)=9.8(0.5-0.3897)=1.08\) m s\(^{-2}\).