Edexcel A-Level Mathematics AS M1.6 Moments Questions

Practise moments in horizontal equilibrium, forming force and reaction equations for rods, beams and planks with particles, supports and tilting limits.

Exam points

form moments about a support to find reactions on a beam or plank
use equal reactions or a breaking tension limit to solve for an unknown mass or position
model tilting about a point by setting the relevant reaction to zero or using equality

Covered syllabus points

M1.6.1 - Moment of a force

[Maximum number: 6]

Figure 2

A beam ADCB has length 5 m. The beam lies on a horizontal step with the end A on the step and the end B projecting over the edge of the step. The edge of the step is at the point D where $D B=1.3 \mathrm{~m}$ , as shown in Figure 2.

When a small boy of mass 30 kg stands on the beam at C, where $C B=0.5 \mathrm{~m}$ , the beam is on the point of tilting.

The boy is modelled as a particle and the beam is modelled as a uniform rod.

(a)

Find the mass of the beam.

A block of mass $X \mathrm{~kg}$ is now placed on the beam at A.
The block is modelled as a particle.

[ 3 ]

$\mathrm{M}(D), m g \times 1.2=30 g \times 0.8$
Other possible equations:

( ↑ ) R=m g+30 g

$\mathrm{M}(A) 2.5 m g+30 g \times 4.5=3.7 R$ $\mathrm{M}(G) 30 g \times 2=1.2 R$

M( C $) \quad m g \times 2=0.8 R$ $\mathrm{M}(B) 2.5 m g+30 g \times 0.5=1.3 R$ $m=20(\mathrm{~kg})$
N.B. Allow an inequality if they state $m=20(\mathrm{~kg})$ at the end

(3)

(b)

Find the smallest value of X that will enable the boy to stand on the beam at B without the beam tilting.

[ 3 ]

$\mathrm{M}(D), X g \times 3.7+20 g \times 1.2=30 g \times 1.3$

N.B. Allow inequality $\geq \ldots$ the correct way round for M1A1ft

M1A1ft

Other possible equations:

( ↑ ) S=m g+30 g+X g

$\mathrm{M}(A) \quad 2.5 m g+30 g \times 5=3.7 S$

where m is their answer from (a).

$\mathrm{M}(G) \quad 30 g \times 2.5=1.2 S+X g \times 2.5$ $\mathrm{M}(B) 2.5 m g+X g \times 5=1.3 S$ $X=\frac{150}{37}, 4.1$ or better $(4.05405 \ldots)$

A1

(3)

[Maximum number: 6]

Figure 1

A non-uniform beam AB has length 6 m and weight W newtons. The beam is supported in equilibrium in a horizontal position by two vertical ropes, one attached to the beam at A and the other attached to the beam at C, where $C B=1.5 \mathrm{~m}$ , as shown in Figure 1.

The centre of mass of the beam is 2.625 m from A.

The ropes are modelled as light strings. The beam is modelled as a non-uniform rod.

Given that the tension in the rope attached at C is 20 N greater than the tension in the rope attached at A,

find the value of W.

$\begin{aligned} T+(T+20)&=W \\ 4.5(T+20)&=2.625W \end{aligned}$

OR another correct pair of equilibrium equations, for example
$\begin{aligned} 2.625T&=1.875(T+20) \\ 4.5T&=1.875W \end{aligned}$

Solve for W.

W=120
Marking guidance for this part:
The A marks and DM1 require using T and T+20, or T and T-20, in both equations.
Award M1 A1 for a correct vertical resolution even if T and T+20 are the wrong way round.
If only $T_A$ and $T_C$ are used, or if $T_A=T_C$ , or if tensions are taken as T and 20 or T and 20T, the maximum is M1 A0 M1 A0 DM0 A0.
If the tension order is unclear, use the candidate's diagram to decide.
W=120 is cao.

[Maximum number: 6]

Figure 2

A beam AB has mass 100 kg and length 8 m. The beam is held in equilibrium in a horizontal position by two vertical ropes attached to the beam at C and D, where $A C=0.5 \mathrm{~m}$ and $B D=0.5 \mathrm{~m}$ . A gymnast of mass $M \mathrm{~kg}$ stands on the beam at the point P, where $A P=2 \mathrm{~m}$ , as shown in Figure 2. The beam remains horizontal and in equilibrium. The tension in the rope attached to the beam at D is 637 N. The gymnast is modelled as a particle, the beam as a uniform rod and the ropes as light inextensible strings.

(a)

Find

[ 6 ]

moments equation
All terms required. Dimensionally correct
Condone sign errors

$M(C), \quad M g \times 1.5+100 g \times 3.5=637 \times 7$ $\mathrm{M}(A): 2 \mathrm{Mg}+400 \mathrm{~g}=\frac{1}{2} T+7.5 \times 637$ $\mathrm{M}(P): 1.5 T+100 \mathrm{~g} \times 2=637 \times 5.5$
Correct unsimplified equation

Resolve vertically or a second moments equation
All terms required. Dimensionally correct
Condone sign errors

$(\uparrow), \quad T+637=M g+100 g$

A1ft

Correct unsimplified equation in M or their M, T or their T

(i)

the value of M,

M=70

A1

$69.99 \ldots=70$ is A0 (from 9.81)

(ii)

the tension in the rope attached to the beam at C.

$T=1000 \mathrm{~N}$ or 1030 N

A1

Not 1029, not 105 g

They need to form two independent equations.

M1A1 for first equation seen

M1A1ft for the second equation (ft on any result from the first eqn) A1 for M, A1 for T.

(6)

(b)

Find the value of x.

[ 6 ]

$2T=60g+100g+48g\Rightarrow T=104g$
Taking moments about C,
60g(1.5)+100g(3.5)+48g(x-0.5)=7T $x=6.5\text{ m}$

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question 3

Question (a)

Question (i)

Question (ii)

Question (b)