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: 10]

Figure 4

A beam AB has mass 30 kg and length 3 m.
The beam rests on supports at C and D where $A C=0.4 \mathrm{~m}$ and $D B=0.4 \mathrm{~m}$ , as shown in Figure 4.

A person of mass 55 kg stands on the beam between C and D.
The person is modelled as a particle at the point P, where C P=x metres and 0<x<2.2

The beam is modelled as a uniform rod resting in equilibrium in a horizontal position.
Using the model,

(a)

show that the magnitude of the reaction at C is (686-245 x) N.

The magnitude of the reaction at C is four times the magnitude of the reaction at D.
Using the model,

[ 3 ]

M(D)
$\left(R_{C} \times 2.2\right)=55 g(2.2-x)+30 g(1.1)$

M1A1

$\left(R_{C}\right)=(686-245 x)(\mathrm{N}) *$

A1 *

N.B. The M mark here is not available if they use $R_{C}=4 R_{D}$ to
obtain the given result.

(3)

(b)

find the value of x

The person steps off the beam and places a package of mass $M \mathrm{~kg}$ at A.
The package is modelled as a particle at the point A.
The beam is now on the point of tilting about C.
Using the model,

[ 4 ]

$R_C=4R_D$ $R_C+R_D=55g+30g,\qquad \frac54(686-245x)=55g+30g$
or a relevant moments equation eliminating $R_C$ and $R_D$ .
x=0.08
(4)

(c)

find the value of M

[ 3 ]

$\mathrm{M}(C):\qquad Mg(0.4)=30g(1.1)$ M=83
or 82.5 or $\frac{165}{2}$ .
(3)

[Maximum number: 5]

Figure 1

Figure 1 shows a beam AB with weight 24 N and length 6 m.
The beam is suspended by two light vertical ropes. The ropes are attached to the points C and D on the beam where A C=x metres and $C D=2 \mathrm{~m}$ .

The tension in the rope attached to the beam at C is double the tension in the rope attached to the beam at D.

The beam is modelled as a uniform rod, resting horizontally in equilibrium.
Find

Figure for Answer to Question 2 — Edexcel A-Level Mathematics AS

Form a moments equation

$\mathrm{M}(A):(2 T \times x)+T(x+2)=(24 \times 3)$

Form a second equation

vert $\quad 3 T=24$

Alternative moments equations in x and T

$\mathrm{M}(C): 24(3-x)=T \times 2$

M(G): 2 T(3-x)=T(x-1)

$\mathrm{M}(D):(2 T \times 2)=24 \times(x-1)$ $\mathrm{M}(B): 2 T(6-x)+T(4-x)=(24 \times 3)$ $\mathrm{M}(C): \frac{24 x}{6} \times \frac{x}{2}+2 T=\frac{24(6-x)}{6} \times \frac{(6-x)}{2}$ $T=8(\mathrm{~N})$ $x=\frac{7}{3}$ accept 2.3 or better

(5)

Notes for question 1

M1

Forms a moments equation in x and T only with the correct no. of terms.

Allow consistent extra g ' s.

M0 if no x.

Correct unsimplified moments equation. Where two moments equations are used, award this mark for the first correct equation.

Resolves vertically to give equation in T only or a second moments equation in x and T (M0 if no x.). Must be dimensionally correct with the correct no. of terms.

Correct value for tension at D

Correct value for x. Accept 2.3 or better

N.B. If T and 2 T the wrong way round or they use 24 g, can score max M1A0M1A0A0.

QUESTION
NUMBER

SCHEME

MARKS

(a)

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

(b)

the value of x.

[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)

All question bank results loaded

Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question 2

Question (a)

Question (b)

Question 3

Question (a)

Question (b)