Moments: past paper questions & worked solutions

Let the tension at D be $T$. The tension at C is double this, so it is $2T$.

The beam is in equilibrium, so the upward forces equal the weight: $2T+T=24$. Hence $3T=24$ and $T=8\ \mathrm{N}$.

The beam is uniform, so its weight acts at the midpoint, $3$ m from A. Also $AC=x$ and $AD=x+2$.

Clockwise and anticlockwise moments balance: $(2T)x+T(x+2)=24\cdot3$.

Substitute $T=8$: $16x+8(x+2)=72$. Thus $24x+16=72$, so $24x=56$ and $x=\dfrac{7}{3}$ m.

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 $51\ \mathrm{N}$.

The rod length is $2a$. The centre of mass is at the midpoint, distance $a$ from A. Also $AC=\dfrac{2a}{5}$ and $AD=2a-\dfrac{3a}{5}=\dfrac{7a}{5}$.

For the largest possible mass at B, the string at C is just slack and the rod is about to tilt about D.

Let the particle mass be $X$. The distance from D to B is $\dfrac{3a}{5}$ and the distance from D to the rod's centre of mass is $\dfrac{2a}{5}$. Hence $Xg\left(\dfrac{3a}{5}\right)=Mg\left(\dfrac{2a}{5}\right)$.

Cancel $ga/5$: $3X=2M$, so $X=\dfrac{2M}{3}$.

Now the particle has mass $\dfrac12M$. Let the tensions at C and D be $T_C$ and $T_D$.

About D, $T_C(a)+\dfrac12Mg\left(\dfrac{3a}{5}\right)=Mg\left(\dfrac{2a}{5}\right)$.

$T_Ca+\dfrac{3}{10}Mga=\dfrac{2}{5}Mga$, so $T_C=\dfrac{1}{10}Mg$.

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

The girder has length $6$ m, with C $1$ m from A and D $1$ m from B. So D is $5$ m from A.

When the $900$ N upward force is applied at A, the girder is about to tilt about D. Taking moments about D gives $900(5)=W(5-x)$.

When the $1500$ N upward force is applied at B, the girder is about to tilt about C. Taking moments about C gives $1500(5)=W(x-1)$.

The equations are $W(5-x)=4500$ and $W(x-1)=7500$. Add them: $W[(5-x)+(x-1)]=12000$, so $4W=12000$ and $W=3000$.

Use $W(x-1)=7500$: $3000(x-1)=7500$, so $x-1=2.5$ and $x=3.5$ m.

From the diagram: ; Distance AC = 2 m ; Distance CD = 8 - 2 - x = (6-x) m ; Distance from A to center = 4 m, so distance from C to center = 4 - 2 = 2 m

$\sum M_C = 0$ Clockwise moments = Anticlockwise moments $90 \times (6-x) = 240 \times 2$ $90(6-x) = 480$ $540 - 90x = 480$ $90x = 60$ $x = \frac{60}{90} = \frac{2}{3}$

$\sum F_{\text{vertical}} = 0$ $T_C + T_D = 240 + W$ $183 + T_D = 240 + W$ $T_D = 57 + W$

$\sum M_C = 0$ Clockwise moments = Anticlockwise moments Distance CD = $6 - \frac{2}{3} = \frac{16}{3}$ m $T_D \times \frac{16}{3} + W \times 2 = 240 \times 2$ $(57 + W) \times \frac{16}{3} + 2W = 480$ $\frac{16(57 + W)}{3} + 2W = 480$ $\frac{912 + 16W}{3} + 2W = 480$ Multiply through by 3: $912 + 16W + 6W = 1440$ $22W = 528$ $W = 24$

The beam length is $5$ m and $DB=1.3$ m, so D is $3.7$ m from A. The centre of mass of the uniform beam is at its midpoint, so it is $1.2$ m to the left of D. Point C is $0.8$ m to the right of D.

When the boy stands at C, the beam is about to tilt about D. Let the mass of the beam be $M$ kg.

$Mg(1.2)=30g(0.8)$. Hence $1.2M=24$ and $M=20\ \mathrm{kg}$.

With the boy at B, his moment about D is $30g(1.3)$. The block at A is $3.7$ m to the left of D, and the beam's weight acts $1.2$ m to the left of D.

$Xg(3.7)+20g(1.2)=30g(1.3)$.

$3.7X+24=39$, so $3.7X=15$ and $X=\dfrac{150}{37}=4.05\ldots\ \mathrm{kg}$.

The block is modelled as a particle, so its mass is treated as acting at one point, A.

Let the tension at A be $T$. The tension at C is then $T+20$. Since $CB=1.5$ m and $AB=6$ m, $AC=4.5$ m.

The upward tensions balance the weight: $T+(T+20)=W$, so $2T+20=W$.

The centre of mass is $2.625$ m from A. Taking moments about A gives $4.5(T+20)=2.625W$.

From the moments equation, $T+20=\dfrac{2.625}{4.5}W=\dfrac{7}{12}W$. Also $T+20=\dfrac{W+20}{2}$. Hence $\dfrac{W+20}{2}=\dfrac{7W}{12}$, so $6W+120=7W$ and $W=120$ N.

Modelling the beam as a rod means it stays straight and rigid, so the distances used in the moments equations do not change.

The branch has length $1.5$ m. Since $AC=0.24$ m and $DB=0.36$ m, $CB=1.26$ m and $CD=0.90$ m. The centre of mass is $x$ m from C.

With the $150$ N upward force at B, the branch is on the point of tilting about C, so the support at D has zero reaction.

The upward force at B balances the weight moment: $150(1.26)=Wx$. Thus $Wx=189$.

With the $225$ N downward force at B, the branch is on the point of tilting about D, so the support at C has zero reaction.

The distance from D to the centre of mass is $0.90-x$. Hence $225(0.36)=W(0.90-x)$.

From the second equation, $81=0.90W-Wx$. Since $Wx=189$, $81=0.90W-189$, so $0.90W=270$ and $W=300\ \mathrm{N}$.

Using $Wx=189$, $300x=189$, so $x=0.63\ \mathrm{m}$.

The beam is $5$ m long. Since $CB=1.5$ m, $AC=3.5$ m. The centre of mass is $1.75$ m from A, so it is also $1.75$ m from C.

For the largest possible $M$ at B, the rope at A is just slack, so the beam is supported at C only.

The particle at B is $1.5$ m to the right of C, and the beam's weight acts $1.75$ m to the left of C. Hence $Mg(1.5)=12g(1.75)$.

$M=\dfrac{12(1.75)}{1.5}=14$.

Let the tension at A be $T$. The tension at C is $2T$. Vertical equilibrium gives $T+2T=12g+15g=27g$, so $T=9g$ and the tension at C is $18g$.

For the new particle at $AD=x$, moments about A give $18g(3.5)=12g(1.75)+15gx$.

Cancel $g$: $63=21+15x$. Thus $15x=42$ and $x=2.8$ m.

The rod is 6 m long and C is 0.4 m from B, so $AC=5.6$ m. The rod is uniform, so its weight $30g$ acts at the midpoint, 3 m from A. The 20 kg particle at B has weight $20g$.

Moments about A remove the tension at A. Clockwise moments from the rod and particle equal the anticlockwise moment from the rope at C: $T_C(5.6)=30g(3)+20g(6)$.

$T_C(5.6)=210g$, so $T_C=37.5g=367.5\ \mathrm{N}$. To suitable accuracy, $T_C=368\ \mathrm{N}$, or $370\ \mathrm{N}$ to 2 significant figures.

Vertically, $T_A+T_C=30g+20g=50g$. Therefore $T_A=50g-37.5g=12.5g=122.5\ \mathrm{N}$, so $T_A=123\ \mathrm{N}$, or $120\ \mathrm{N}$ to 2 significant figures.

For the greatest possible mass at B, the rope at A is just slack, so $T_A=0$. Take moments about C: the clockwise moment of $Mg$ at B balances the anticlockwise moment of the rod weight.

The distance from C to B is $0.4$ m and the rod weight acts $5.6-3=2.6$ m to the left of C. Hence $Mg(0.4)=30g(2.6)$, so $M=\dfrac{78}{0.4}=195$.

Vertically, $R_C+R_D=50g+25g=75g$. Given $2R_D=3R_C$, we have $R_D=1.5R_C$. Therefore $2.5R_C=75g$, so $R_C=30g$ and $R_D=45g$.

$D$ is $6-1.8=4.2$ m from A. The support C is $4.2-0.9=3.3$ m to the left of D, the child at E is $4.2-3=1.2$ m to the left of D, and the centre of mass is $x$ m to the left of D.

Balancing moments about D gives $50gx+25g(1.2)=R_C(3.3)$.

Substitute $R_C=30g$: $50x+30=99$, so $50x=69$ and $x=\dfrac{69}{50}=1.38$.

When the block at B is on the point of making the beam tilt, the beam is about to pivot about D and the reaction at C is zero.

The child and the beam act to the left of D, and the block at B acts $1.8$ m to the right of D. Hence $25g(1.2)+50g(1.38)=Mg(1.8)$.

$30+69=1.8M$, so $M=\dfrac{99}{1.8}=55$.

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 $0.625\le M\le16.25$.

Take moments about A. The rope at C is $2.1$ m from A and the weight of the uniform rod acts at its midpoint, $1.5$ m from A. Hence $350(2.1)=W(1.5)$.

$735=1.5W$, so $W=490\ \mathrm{N}$.

When P is attached, the tension at C is $600$ N. Taking moments about A: $600(2.1)=490(1.5)+210d$.

$1260=735+210d$, so $525=210d$ and $d=2.5$.

When Q is attached at B and the rod is on the point of tilting, the rope at A is just slack. The rod is about to tilt about C.

Take moments about C. The rod weight acts $2.1-1.5=0.6$ m to the left of C, and Q acts $3-2.1=0.9$ m to the right. Thus $490(0.6)=X(0.9)$, so $X=\dfrac{980}{3}\ \mathrm{N}$.

$AC=0.4$ 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 $M=\dfrac{33}{0.4}=82.5$.

The beam length is $8a$. C is at distance $a$ from A, and D is at distance $6a$ from A because $DB=2a$.

Let the reaction at C be $R$. The reaction at D is $3R$.

$R+3R=60g$, so $4R=60g$ and $R=15g$.

The beam's weight acts at distance $x$ from A. Hence $60g x=R(a)+3R(6a)$.

Substitute $R=15g$: $60g x=15ga+45g(6a)=285ga$. Thus $x=\dfrac{285a}{60}=\dfrac{19a}{4}$.

When the box is added, let each support reaction be $S$. Vertical equilibrium gives $2S=(60+M)g$.

The box is at $AE=2a$. Moments about A give $60g\left(\dfrac{19a}{4}\right)+Mg(2a)=S(a)+S(6a)=7aS$.

Substitute $S=\dfrac{(60+M)g}{2}$ and cancel $ag$: $285+2M=\dfrac{7}{2}(60+M)=210+\dfrac72M$. Hence $75=\dfrac32M$ and $M=50$.

The beam is $2$ m long. Since $AC=0.6$ m and $DB=0.2$ m, point D is $1.8$ m from A. The uniform beam has weight $mg$ acting $1.0$ m from A.

Let the tension at D be $T$. The tension at C is four times this, so it is $4T$.

$4T+T=pmg+mg$, so $5T=(p+1)mg$.

The particle at A has no moment about A. Hence $4T(0.6)+T(1.8)=mg(1.0)$.

$2.4T+1.8T=mg$, so $4.2T=mg$ and $T=\dfrac{5mg}{21}$.

Substitute into $5T=(p+1)mg$: $5\cdot\dfrac{5mg}{21}=(p+1)mg$. Thus $p+1=\dfrac{25}{21}$ and $p=\dfrac{4}{21}$.

When the particle of mass $qm$ is at A and the beam is on the point of tilting, the rope at D is just slack, so its tension is zero.

About C, the particle at A has moment $qmg(0.6)$ and the beam weight has moment $mg(0.4)$ in the opposite direction. Thus $qmg(0.6)=mg(0.4)$, giving $q=\dfrac{2}{3}$.

The beam is uniform, so its centre of mass is at the midpoint of the beam and its weight acts there.