Pearson Edexcel IAL Mathematics M1.6.1 Moment of a force

Practise taking moments of coplanar parallel forces on rods and beams, using equilibrium to find reactions, masses, positions and limiting loads.

[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 ]

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

M1 A1

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 $) \quad m g \times 2=0.8 R$ $\mathrm{M}(B) 2.5 m g+30 g \times 0.5=1.3 R$ $m=20(\mathrm{~kg})$

A1

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)

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M1.6.1 - Moment of a force question 1

Question (a)

Question (b)