Forces and Friction: past paper questions & worked solutions

Take east as $\mathbf{i}$ and north as $\mathbf{j}$. The force due west is $-10\mathbf{i}$. A bearing of $330^\circ$ is $30^\circ$ west of north, so the 8 N force has components $-8\sin30^\circ\mathbf{i}+8\cos30^\circ\mathbf{j}$.

$\mathbf{F}=(-10-4)\mathbf{i}+4\sqrt3\mathbf{j}=-14\mathbf{i}+4\sqrt3\mathbf{j}$.

$|\mathbf{F}|=\sqrt{(-14)^2+(4\sqrt3)^2}=\sqrt{196+48}=\sqrt{244}=2\sqrt{61}\ \mathrm{N}$.

$2\sqrt{61}=15.620\ldots$, so the magnitude is about $16\ \mathrm{N}$ to 2 significant figures.

Taking moments about A: $\sum M_A = 0$ Clockwise moments = Anticlockwise moments $P \times 3a = W \times d$ $3aP = Wd \quad \text{...(1)}$

Taking moments about C: $\sum M_C = 0$ Distance from C to COM = $(2a - d)$ Distance from C to B = $a$ $1.25P \times a = W \times (2a - d)$ $1.25Pa = W(2a - d)$ $1.25Pa = 2aW - Wd \quad \text{...(2)}$

From (1): $Wd = 3aP$ $1.25Pa = 2aW - 3aP$ $1.25Pa + 3aP = 2aW$ $4.25Pa = 2aW$ $P = \frac{2aW}{4.25a}$ $P = \frac{2W}{4.25}$ Substitute back into (1): $3a \times \frac{2W}{4.25} = Wd$ $\frac{6aW}{4.25} = Wd$ $d = \frac{6a}{4.25}$ $d = \frac{6a}{17/4}$ $d = \frac{24a}{17}$ $d = 1.41a \text{ (3 sf)}$ Or in exact form: $d = \frac{24a}{17}$

The rod is 6.5 m long and overhangs C by 1.5 m, so C is 5.0 m from A. The rod centre of mass is 3.0 m from A, so it is 2.0 m to the left of C. Point D is 0.5 m from A, so it is 4.5 m to the left of C.

For the smallest $M$, the rod is just prevented from tilting clockwise about C. Taking moments about C: $Mg(4.5)+1.2g(2)=4g(1.5)$.

$4.5M+2.4=6$, so $4.5M=3.6$ and $M=0.8$.

Now the old particles are removed. Since $EB=0.9$ m and $AB=6.5$ m, E is $5.6$ m from A, so E is $0.6$ m to the right of C.

At tilting about C, moments about C give $Xg(0.6)=1.2g(2)$, so $X=4$.

The upward force at C balances the rod and the new particle: $R_C=1.2g+4g=5.2g\ \mathrm{N}$, which is about $51\ \mathrm{N}$.

Given $\tan\theta=\dfrac34$, use $\sin\theta=\dfrac35$ and $\cos\theta=\dfrac45$.

The horizontal force of $2$ N is balanced by the horizontal component of the tension in the sloping part of the string: $T\cos\theta=2$.

$T\cdot\dfrac45=2$, so $T=2.5\ \mathrm{N}$.

The ring is smooth, so the tension is the same in both parts of the string. Upward components are $T$ from the vertical part and $T\sin\theta$ from the sloping part. Therefore $T+T\sin\theta=Mg$.

$Mg=2.5+2.5\cdot\dfrac35=4.0$, so $M=\dfrac{4.0}{9.8}=0.408\ \mathrm{kg}$.

The force $F$ is perpendicular to the string. The weight is $5$ N vertically downward, and the string makes $60^\circ$ with the vertical.

The component of the weight perpendicular to the string is $5\cos30^\circ$, so $F=5\cos30^\circ=\dfrac{5\sqrt3}{2}\ \mathrm{N}$.

The tension balances the component of the weight along the string. This component is $5\sin30^\circ$.

$T=5\sin30^\circ=\dfrac52\ \mathrm{N}$.

C is $2.5$ m from A. Since $DB=2$ m and $AB=9$ m, D is $7$ m from A.

Let the reaction at C be $R$. The reaction at D is twice this, so it is $2R$.

$R+2R=Mg$, so $3R=Mg$.

$R(2.5)+2R(7)=Mgd$.

$2.5R+14R=3Rd$. Hence $16.5R=3Rd$.

Cancel $R$: $d=\dfrac{16.5}{3}=5.5\ \mathrm{m}$.

$\sum F_{\perp} = 0 \text{ (no sideways motion)}$ Perpendicular component of 800 N rope: $800 \sin 25° = 800 \times 0.4226 = 338.1 \text{ N}$ Perpendicular component of 500 N rope: $500 \sin \alpha$

$800 \sin 25° = 500 \sin \alpha$ $338.1 = 500 \sin \alpha$ $\sin \alpha = \frac{338.1}{500}$ $\sin \alpha = 0.6762$ $\alpha = \sin^{-1}(0.6762)$ $\alpha = 42.5°$

$\sum F_{\parallel} = ma$ Parallel component of 800 N rope: $800 \cos 25° = 800 \times 0.9063 = 725.0 \text{ N}$ Parallel component of 500 N rope: $500 \cos 42.5° = 500 \times 0.7373 = 368.7 \text{ N}$ Resistance force: $750$ N (opposes motion) Mass: $15$ tonnes $= 15000$ kg

$725.0 + 368.7 - 750 = 15000 \times a$ $343.7 = 15000a$ $a = \frac{343.7}{15000}$ $a = 0.02291...$ $a = 0.023 \text{ m s}^{-2} \text{ (2 sf)}$

Given $\tan\alpha=\dfrac43$, use $\sin\alpha=\dfrac45$ and $\cos\alpha=\dfrac35$.

The applied force increases the normal reaction in this model, so $R=W+P\sin\alpha$.

For the particle not to move, friction must be able to balance the horizontal component of the applied force, so $P\cos\alpha\le F_{\max}$.

Since $\mu=\dfrac14$, $F_{\max}=\dfrac14R=\dfrac14(W+P\sin\alpha)$.

Thus $P\cdot\dfrac35\le\dfrac14\left(W+P\cdot\dfrac45\right)$.

Multiplying by $20$ gives $12P\le5W+4P$. Hence $8P\le5W$, so $P\le\dfrac58W$.

A bearing of $240^\circ$ is $60^\circ$ west of south. Hence $\mathbf{F}=(-8\sin60^\circ)\mathbf{i}+(-8\cos60^\circ)\mathbf{j}=-4\sqrt3\mathbf{i}-4\mathbf{j}$.

The force $\mathbf{G}$ is due south with magnitude $10$, so $\mathbf{G}=-10\mathbf{j}$. Therefore $\mathbf{R}=\mathbf{F}+\mathbf{G}=-4\sqrt3\mathbf{i}-14\mathbf{j}$.

$|\mathbf{R}|=\sqrt{(-4\sqrt3)^2+(-14)^2}=\sqrt{48+196}=\sqrt{244}\ \mathrm{N}$.

The resultant points south-west. The angle west of south satisfies $\tan\theta=\dfrac{4\sqrt3}{14}=\dfrac{2\sqrt3}{7}$.

$\theta=26.3^\circ$. Bearing $=180^\circ+26.3^\circ=206.3^\circ$, so the bearing is $206^\circ$ to the nearest degree.

Since $\mathbf F$ has magnitude $4$ N on a bearing of $120^\circ$, $2\mathbf F$ has magnitude $8$ N on the same bearing.

$2\mathbf F=8\sin120^\circ\mathbf i+8\cos120^\circ\mathbf j=4\sqrt3\mathbf i-4\mathbf j$. Also $\mathbf G=6\mathbf j$.

$\mathbf P=2\mathbf F+\mathbf G=4\sqrt3\mathbf i+2\mathbf j$.

$|\mathbf P|=\sqrt{(4\sqrt3)^2+2^2}=\sqrt{48+4}=\sqrt{52}$ N.

The vector is in the north-east quadrant. The angle east of north satisfies $\tan\theta=\dfrac{4\sqrt3}{2}=2\sqrt3$.

$\theta=73.9^\circ$ approximately, so the bearing is $074^\circ$ to the nearest degree.

The resultant force is $(2p\mathbf{i}-3p\mathbf{j})+(q\mathbf{i}+\mathbf{j})+(3\mathbf{i}-q\mathbf{j})=(2p+q+3)\mathbf{i}+(-3p+1-q)\mathbf{j}$.

The resultant acts in the direction $4\mathbf{i}-5\mathbf{j}$, so the ratio of the components is the same: $\dfrac{-3p+1-q}{2p+q+3}=\dfrac{-5}{4}$.

Cross-multiplying gives $4(-3p+1-q)=-5(2p+q+3)$. Therefore $-12p+4-4q=-10p-5q-15$, so $2p-q=19$.

With $p=7$, the relation gives $14-q=19$, hence $q=-5$.

Substitute $p=7$ and $q=-5$ into the resultant: $\mathbf{F}=12\mathbf{i}-15\mathbf{j}$. Since the mass is $0.5$ kg, $\mathbf{F}=m\mathbf{a}$ gives $12\mathbf{i}-15\mathbf{j}=0.5\mathbf{a}$.

Dividing by $0.5$ gives $\mathbf{a}=24\mathbf{i}-30\mathbf{j}\ \mathrm{m\,s^{-2}}$.

$\mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3=(6\mathbf{i}+8\mathbf{j})+(-16\mathbf{i}+2\mathbf{j})+(-2\mathbf{i}+8\mathbf{j})=-12\mathbf{i}+18\mathbf{j}$.

The mass is $2.5$ kg, so $-12\mathbf{i}+18\mathbf{j}=2.5\mathbf{a}$. Hence $\mathbf{a}=-4.8\mathbf{i}+7.2\mathbf{j}\ \mathrm{m\,s^{-2}}$.

$|\mathbf{a}|=\sqrt{(-4.8)^2+7.2^2}=8.653\ldots$, so the magnitude is $8.7\ \mathrm{m\,s^{-2}}$ to 2 significant figures. Exactly, $|\mathbf{a}|=\dfrac{12\sqrt{13}}{5}$.

With $\mathbf{F}_4=p\mathbf{i}+p\mathbf{j}$, the new resultant is $(p-12)\mathbf{i}+(p+18)\mathbf{j}$.

The resultant acts in the direction $7\mathbf{i}+2\mathbf{j}$, so $\dfrac{p-12}{p+18}=\dfrac{7}{2}$.

$2(p-12)=7(p+18)$, so $2p-24=7p+126$. Therefore $-150=5p$ and $p=-30$.

The two given forces have angle $150^\circ$ between them, so the triangle used for adding them has the supplementary angle $30^\circ$ between the side of length $X$ and the side of length $5\sqrt3$.

The resultant has magnitude $\sqrt{129}$. Therefore $129=X^2+(5\sqrt3)^2-2X(5\sqrt3)\cos30^\circ$.

Since $(5\sqrt3)^2=75$ and $2(5\sqrt3)\cos30^\circ=15$, the equation becomes $129=X^2+75-15X$. Thus $X^2-15X-54=0$, giving $X=18$.

Let $\beta$ be the angle between Q and the resultant. Using the cosine rule in the same triangle, $\dfrac{18}{\sin\beta}=\dfrac{\sqrt{129}}{\sin30^\circ}$.

This gives an obtuse angle for $\beta$. Equivalently, $\cos\beta=\dfrac{(5\sqrt3)^2+129-18^2}{2(5\sqrt3)\sqrt{129}}$, so $\beta=127.6\ldots^\circ$.

To the nearest degree, the angle between Q and the resultant is $128^\circ$.

The boat travels straight, so the sideways components balance: $500\sin40^\circ=P\sin\alpha$.

The speed is constant, so the forward components balance the resistance: $500\cos40^\circ+P\cos\alpha=900$.

From the two equations, $P\sin\alpha=500\sin40^\circ$ and $P\cos\alpha=900-500\cos40^\circ$.

Divide the sine equation by the cosine equation: $\tan\alpha=\dfrac{500\sin40^\circ}{900-500\cos40^\circ}$. This gives $\alpha=31.9^\circ$, so $\alpha=32^\circ$ approximately.

Use $P\sin\alpha=500\sin40^\circ$: $P=\dfrac{500\sin40^\circ}{\sin31.9^\circ}=608.7\ldots\ \mathrm{N}$. Hence $P\approx610\ \mathrm{N}$.

Since $\tan\alpha=\dfrac34$, $\sin\alpha=\dfrac35$ and $\cos\alpha=\dfrac45$.

The force $X$ acts along the plane, so the normal reaction is $R=mg\cos\alpha=\dfrac45mg$.

The limiting friction is $F=\mu R=\dfrac45\mu mg$.

When $X=2P$, the particle is about to slide up, so friction acts down the plane. Hence $2P=mg\sin\alpha+F=\dfrac35mg+\dfrac45\mu mg$.

When $X=P$, the particle is about to slide down, so friction acts up the plane. Hence $P+F=mg\sin\alpha$, so $P=\dfrac35mg-\dfrac45\mu mg$.

Double the second equation: $2P=\dfrac65mg-\dfrac85\mu mg$. This equals the first expression for $2P$.

$\dfrac65mg-\dfrac85\mu mg=\dfrac35mg+\dfrac45\mu mg$. Cancel $mg$ and multiply by $5$: $6-8\mu=3+4\mu$. Hence $3=12\mu$ and $\mu=\dfrac14$.

Take distances from A. Then C is at $1$ m, D is at $3$ m, the rod's centre of mass is at $2.5$ m, and B is at $5$ m.

Taking moments about D: $R_C(2)+Mg(2)=10g(3)+5g(0.5)$.

$2R_C+2Mg=30g+2.5g=32.5g$, so $R_C=16.25g-Mg=(16.25-M)g$.

Taking moments about C: $10g(1)+R_D(2)=5g(1.5)+Mg(4)$.

$10g+2R_D=7.5g+4Mg$, so $2R_D=(4M-2.5)g$ and $R_D=(2M-1.25)g$.

For the rod to rest on both supports, both reactions must be non-negative.

$R_C\ge0$ gives $16.25-M\ge0$, so $M\le16.25$. Also $R_D\ge0$ gives $2M-1.25\ge0$, so $M\ge0.625$. Therefore $0.625\le M\le16.25$.

Equilibrium gives $\mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3=\mathbf{0}$. Now $\mathbf{F}_1+\mathbf{F}_2=(5\mathbf{i}+2\mathbf{j})+(-3\mathbf{i}+\mathbf{j})=2\mathbf{i}+3\mathbf{j}$.

So $\mathbf{F}_3=-2\mathbf{i}-3\mathbf{j}$.

The vector $-\mathbf{j}$ points vertically downward. For $\mathbf{F}_3=-2\mathbf{i}-3\mathbf{j}$, the angle from $-\mathbf{j}$ satisfies $\tan\theta=\dfrac{2}{3}$.

$\theta=33.7^\circ$, so the angle is about $34^\circ$.

With $\mathbf{F}_4=\lambda(\mathbf{i}+3\mathbf{j})$, the resultant is $(2+\lambda)\mathbf{i}+(3+3\lambda)\mathbf{j}$.

The mass is $4$ kg and acceleration magnitude is $3.25$, so the resultant force magnitude is $4(3.25)=13$ N.

$(2+\lambda)^2+(3+3\lambda)^2=13^2$.

Expanding gives $10\lambda^2+22\lambda-156=0$. The roots are $3$ and $-5.2$, and since $\lambda$ is positive, $\lambda=3$.

For vertical equilibrium, $R+200\sin15^\circ+T\sin25^\circ=20g$.

For the smallest $T$, friction is limiting and balances the remaining horizontal tendency. Horizontally, $200\cos15^\circ-T\cos25^\circ-F=0$.

$F=0.3R$.

From the vertical equation, $R=20g-200\sin15^\circ-T\sin25^\circ$. Substitute into $F=0.3R$ and then into the horizontal equation.

$200\cos15^\circ-T\cos25^\circ-0.3(20g-200\sin15^\circ-T\sin25^\circ)=0$. Solving gives $T=192.3\ldots\ \mathrm{N}$, so the smallest value is about $192\ \mathrm{N}$.

$AC=0.4$ m and $DB=0.4$ m, so $CD=2.2$ m. The beam is uniform, so its weight $30g$ acts at the midpoint, which is $1.1$ m from D. The person is $2.2-x$ m from D.

Taking moments about D removes $R_D$: $R_C(2.2)=55g(2.2-x)+30g(1.1)$.

Using $g=9.8$, $R_C=\dfrac{539(2.2-x)+294(1.1)}{2.2}=686-245x$.

Given $R_C=4R_D$ and vertical equilibrium gives $R_C+R_D=(55+30)g=833$. Hence $5R_D=833$ and $R_C=666.4$.

$686-245x=666.4$, so $245x=19.6$ and $x=0.08$.

When the package is at A and the beam is on the point of tilting about C, the reaction at D is zero.

Take moments about C: $Mg(0.4)=30g(1.1)$. Thus $M=\dfrac{33}{0.4}=82.5$.

For equilibrium, $\mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3=\mathbf{0}$. Hence $\mathbf{F}_3=-(\mathbf{F}_1+\mathbf{F}_2)$.

$\mathbf{F}_1+\mathbf{F}_2=(3c-14)\mathbf{i}+(4c+7)\mathbf{j}$, so $\mathbf{F}_3=(14-3c)\mathbf{i}+(-7-4c)\mathbf{j}$.

When $c=2$, $\mathbf{F}_1=6\mathbf{i}+8\mathbf{j}$ and $\mathbf{F}_2=-14\mathbf{i}+7\mathbf{j}$. Therefore the resultant is $-8\mathbf{i}+15\mathbf{j}$.

The resultant is in the second quadrant. Its angle from the positive $\mathbf{i}$ direction is $180^\circ-\tan^{-1}\left(\dfrac{15}{8}\right)=118.1^\circ$, so the angle is about $118^\circ$.

$|\mathbf{R}|=\sqrt{(-8)^2+15^2}=17\ \mathrm{N}$.

The acceleration magnitude is $8.5\ \mathrm{m\,s^{-2}}$, so $17=8.5m$. Hence $m=2$.

Since $\tan\alpha=\dfrac43$, $\sin\alpha=\dfrac45$ and $\cos\alpha=\dfrac35$.

The vertical component of the thrust is $2.5\cos\alpha=2.5\cdot\dfrac35=1.5$ N upward. The weight is $0.2g=1.96$ N downward.

For equilibrium, friction must supply the missing upward force: $F=1.96-1.5=0.46\ \mathrm{N}$. So friction acts upwards.

When $T=6.125$, the vertical component of the thrust is larger than the weight, so friction acts downward at the limiting value.

The normal reaction is the horizontal component of the thrust: $R=6.125\sin\alpha=6.125\cdot\dfrac45=4.9\ \mathrm{N}$.

Vertical equilibrium gives $0.2g+F=6.125\cos\alpha$. Thus $1.96+F=6.125\cdot\dfrac35=3.675$, so $F=1.715$ N.

$F=\mu R$, so $\mu=\dfrac{1.715}{4.9}=0.35$.

The two tensions are equal and symmetric. Their vertical components support the weight of C, so $2T\sin\beta=3mg$. Hence $T=\dfrac{3mg}{2\sin\beta}$.

For ring A, the normal reaction balances its own weight and the downward vertical component of the string tension: $R=mg+T\sin\beta$.

From $2T\sin\beta=3mg$, we have $T\sin\beta=\dfrac32mg$. Therefore $R=mg+\dfrac32mg=\dfrac52mg$.

The rings are just about to slide towards each other, so friction is limiting. With $\mu=\dfrac45$, $F=\mu R=\dfrac45\cdot\dfrac52mg=2mg$.

The horizontal component of the tension is balanced by friction: $T\cos\beta=F=2mg$.

Substitute $T=\dfrac{3mg}{2\sin\beta}$: $\dfrac{3mg}{2\sin\beta}\cos\beta=2mg$. Hence $\dfrac32\cot\beta=2$, so $\cot\beta=\dfrac43$ and $\tan\beta=\dfrac34$.

Since $\tan\alpha=\dfrac34$, $\sin\alpha=\dfrac35$ and $\cos\alpha=\dfrac45$.

If there is no friction, the component of the horizontal force up the plane balances the component of weight down the plane: $X\cos\alpha=2g\sin\alpha$.

$14.7\cdot\dfrac45=11.76$, and $2(9.8)\cdot\dfrac35=11.76$. The components balance exactly, so the frictional force is zero.

For the smallest possible $X$, the particle is about to slide down the plane, so friction acts up the plane.

The horizontal force increases the normal reaction, so $R=2g\cos\alpha+X\sin\alpha$.

With $\mu=0.5$, the limiting friction is $F=0.5R=0.5(2g\cos\alpha+X\sin\alpha)$.

At the limiting case, $X\cos\alpha+F=2g\sin\alpha$.

Substitute the exact trig values: $\dfrac45X+0.5\left(2g\cdot\dfrac45+X\cdot\dfrac35\right)=2g\cdot\dfrac35$. This gives $\dfrac{11}{10}X+\dfrac{4g}{5}=\dfrac{6g}{5}$, so $X=\dfrac{4g}{11}=3.56\ldots\ \mathrm{N}$.

P is on the point of moving up the plane, so friction acts down the plane. The tension has component $0.75mg\cos\alpha$ up the plane, and the weight component is $mg\sin\alpha$ down the plane.

For equilibrium parallel to the plane, $0.75mg\cos\alpha=mg\sin\alpha+F$. Hence $F=0.75mg\cos\alpha-mg\sin\alpha$.

The string has a component away from the plane, so $R=mg\cos\alpha-0.75mg\sin\alpha$.

Given $\mu=\dfrac12$, limiting friction gives $F=\dfrac12R$. Therefore $0.75mg\cos\alpha-mg\sin\alpha=\dfrac12(mg\cos\alpha-0.75mg\sin\alpha)$.

Divide by $mg$ and collect terms: $0.75\cos\alpha-\sin\alpha=0.5\cos\alpha-0.375\sin\alpha$. Thus $0.25\cos\alpha=0.625\sin\alpha$, so $\tan\alpha=\dfrac{0.25}{0.625}=\dfrac25$.

Without the string, $R=mg\cos\alpha$ and the maximum friction is $\dfrac12mg\cos\alpha$. The component of weight down the plane is $mg\sin\alpha$.

Since $\tan\alpha=\dfrac25<\dfrac12$, we have $mg\sin\alpha<\dfrac12mg\cos\alpha$. The available friction is large enough to hold P, so P remains at rest.

The string makes angle $\theta$ with the vertical rod and $\tan\theta=\dfrac{12}{5}$. Hence $\sin\theta=\dfrac{12}{13}$ and $\cos\theta=\dfrac{5}{13}$.

The upward component of the tension is $10\cos\theta=10\cdot\dfrac{5}{13}=\dfrac{50}{13}$ N. The weight is $0.2g=1.96$ N.

Since the upward component of tension is larger than the weight, friction must act downwards. Its magnitude is $\dfrac{50}{13}-1.96=1.89\ldots\ \mathrm{N}$, so $F=1.9\ \mathrm{N}$.

For variable tension $T$, the normal reaction from the rod is produced by the horizontal component of tension: $R=T\sin\theta=\dfrac{12T}{13}$.

Since $\mu=\dfrac14$, limiting friction is $F=\mu R=\dfrac14\cdot\dfrac{12T}{13}=\dfrac{3T}{13}$.

For the minimum tension, the ring would tend to slide down, so friction acts upwards. Vertical equilibrium gives $T\cos\theta+F=0.2g$, hence $\dfrac{5T}{13}+\dfrac{3T}{13}=1.96$. Thus $T=3.185\ldots\ \mathrm{N}$, so $T_{\min}=3.2\ \mathrm{N}$.

For the maximum tension, the ring would tend to move up, so friction acts downwards. Vertical equilibrium gives $T\cos\theta=0.2g+F$, hence $\dfrac{5T}{13}=1.96+\dfrac{3T}{13}$. Thus $T=12.74\ldots\ \mathrm{N}$, so $T_{\max}=13\ \mathrm{N}$.

Resolve vertically. The upward component of the applied force is $14\sin30^\circ$, so $R+14\sin30^\circ=5g$.

$R+7=49$, so $R=42\ \mathrm{N}$.

$F_{\max}=\mu R=\dfrac37(42)=18\ \mathrm{N}$.

The horizontal component of the applied force is $14\cos30^\circ=12.1\ldots\ \mathrm{N}$. Since this is less than the maximum possible friction, the block does not move.

For equilibrium, the actual friction equals the horizontal pull, so $F=12.1\ \mathrm{N}$ approximately.

When the block is on the point of slipping, friction is limiting. Resolve vertically: $R+P\sin30^\circ=5g$, so $R=5g-P\sin30^\circ$.

At the limiting case, $P\cos30^\circ=\dfrac37R=\dfrac37(5g-P\sin30^\circ)$.

Substitute $g=9.8$, $\sin30^\circ=\dfrac12$: $P\cos30^\circ=\dfrac37(49-\dfrac12P)$. Solving gives $P=19.4\ldots\ \mathrm{N}$.

The particle is on the point of sliding down the plane, so friction acts up the plane. Also $X=mg$.

Both the weight $kmg$ and the horizontal force $mg$ press the particle into the plane. Therefore $R=kmg\cos\theta+mg\sin\theta$.

Down the plane is the component $kmg\sin\theta$. Up the plane are friction $F$ and the component $mg\cos\theta$. Thus $F+mg\cos\theta=kmg\sin\theta$.

So $F=kmg\sin\theta-mg\cos\theta$.

$\mu=\dfrac{F}{R}=\dfrac{kmg\sin\theta-mg\cos\theta}{kmg\cos\theta+mg\sin\theta}$.

Cancel $mg$ and divide numerator and denominator by $\cos\theta$: $\mu=\dfrac{k\tan\theta-1}{k+\tan\theta}$, as required.

When $k=1$, $\mu=\dfrac{\tan\theta-1}{1+\tan\theta}$. Since $\mu>0$ and the denominator is positive, $\tan\theta-1>0$.

Therefore $\tan\theta>1$, so because $\theta$ is acute, $\theta>45^\circ$.

$\mathbf{F}_{\text{resultant}} = \mathbf{F}_1 + \mathbf{F}_2$ $\mathbf{F}_{\text{resultant}} = (-2\mathbf{i} + 3\mathbf{j}) + (4\mathbf{i} + 2\mathbf{j})$ $= (-2 + 4)\mathbf{i} + (3 + 2)\mathbf{j}$ $= 2\mathbf{i} + 5\mathbf{j} \text{ N}$

$\mathbf{F} = m\mathbf{a}$ $2\mathbf{i} + 5\mathbf{j} = 2 \times \mathbf{a}$ $\mathbf{a} = \frac{2\mathbf{i} + 5\mathbf{j}}{2}$ $\mathbf{a} = \mathbf{i} + 2.5\mathbf{j} \text{ m s}^{-2}$ Or: $\mathbf{a} = \mathbf{i} + \frac{5}{2}\mathbf{j}$ m s$^{-2}$

$\mathbf{v} = \mathbf{u} + \mathbf{a}t$ Given: ; $\mathbf{u} = 3\mathbf{i} - 4\mathbf{j}$ m s$^{-1}$ ; $\mathbf{a} = \mathbf{i} + 2.5\mathbf{j}$ m s$^{-2}$ ; $t = 3$ seconds $\mathbf{v} = (3\mathbf{i} - 4\mathbf{j}) + (\mathbf{i} + 2.5\mathbf{j}) \times 3$ $= 3\mathbf{i} - 4\mathbf{j} + 3\mathbf{i} + 7.5\mathbf{j}$ $= 6\mathbf{i} + 3.5\mathbf{j} \text{ m s}^{-1}$

$|\mathbf{v}| = \sqrt{v_x^2 + v_y^2}$ $\text{Speed} = \sqrt{6^2 + 3.5^2}$ $= \sqrt{36 + 12.25}$ $= \sqrt{48.25}$ $= 6.946...$ $= 6.95 \text{ m s}^{-1} \text{ (3 s.f.)}$

$\mathbf{F}_{\text{resultant}} = \mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3$ $= (-2\mathbf{i} + 3\mathbf{j}) + (4\mathbf{i} + 2\mathbf{j}) + (b\mathbf{i} + c\mathbf{j})$ $= (-2 + 4 + b)\mathbf{i} + (3 + 2 + c)\mathbf{j}$ $= (2 + b)\mathbf{i} + (5 + c)\mathbf{j}$

Given that resultant = $\lambda(\mathbf{i} + \mathbf{j})$, we can write: $(2 + b)\mathbf{i} + (5 + c)\mathbf{j} = \lambda\mathbf{i} + \lambda\mathbf{j}$

For this to be in the form $\lambda(\mathbf{i} + \mathbf{j})$, the coefficients must be equal: $2 + b = \lambda$ $5 + c = \lambda$ Therefore: $2 + b = 5 + c$ $b = 5 + c - 2$ $b = 3 + c$ $b - c = 3$

Resultant = $(2 + b)\mathbf{i} + (5 + c)\mathbf{j}$ Since resultant = $\lambda(\mathbf{i} + \mathbf{j})$, we have $2 + b = 5 + c = \lambda$ $|\mathbf{F}_{\text{resultant}}| = \lambda\sqrt{1^2 + 1^2} = \lambda\sqrt{2}$ Given: $|\mathbf{F}_{\text{resultant}}| = 10\sqrt{2}$ $\lambda\sqrt{2} = 10\sqrt{2}$ $\lambda = 10 \quad \text{or} \quad \lambda = -10$ (The negative value comes from considering the opposite direction)

Case 1: $\lambda = 10$ $2 + b = 10 \quad \Rightarrow \quad b = 8$ $5 + c = 10 \quad \Rightarrow \quad c = 5$ Check: $b - c = 8 - 5 = 3$ ✓ Therefore: $\mathbf{F}_3 = 8\mathbf{i} + 5\mathbf{j}$ N Case 2: $\lambda = -10$ $2 + b = -10 \quad \Rightarrow \quad b = -12$ $5 + c = -10 \quad \Rightarrow \quad c = -15$ Check: $b - c = -12 - (-15) = 3$ ✓ Therefore: $\mathbf{F}_3 = -12\mathbf{i} - 15\mathbf{j}$ N

Before B hits the floor, it descends $1.4$ m from rest in $2$ s. Use $s=ut+\dfrac12at^2$: $1.4=\dfrac12a(2)^2$. Hence $a=0.7\ \mathrm{m\,s^{-2}}$.

The string is inextensible, so A and B have the same magnitude of acceleration while the string is taut.

For B, taking downward as positive: $3g-T=3(0.7)$. Thus $T=3g-2.1=27.3$ N.

The pulley is pulled by two perpendicular tensions of magnitude $T$. Therefore the resultant is $\sqrt{T^2+T^2}=T\sqrt2=27.3\sqrt2=38.6$ N approximately.

For A before B hits the floor, $T-F=4(0.7)$. Hence $F=27.3-2.8=24.5$ N.

On the horizontal table, $R=4g=39.2$ N and $F=\mu R$. So $\mu=\dfrac{24.5}{39.2}=0.625$.

At this instant, $v=at=0.7(2)=1.4\ \mathrm{m\,s^{-1}}$. A has also moved $1.4$ m towards the pulley.

After B hits the floor, A is slowed only by friction. Its deceleration is $\mu g=0.625(9.8)=6.125\ \mathrm{m\,s^{-2}}$. Use $0=(1.4)^2-2(6.125)s$, giving $s=0.16$ m.

A's total distance is $1.4+0.16=1.56$ m, which is less than $2$ m. Therefore A does not reach the pulley.