Question

A uniform wooden plank of
mass 30 kg is supported symmetrically by two
light identical cables; each can sustain a tension
up to 500 N. After tying, the cables are exactly
vertical and are separated by 2 m. A boy of mass
50 kg, standing at the centre of the plank, is
interested in walking on the plank. How far can
he walk? (g= 10 ms 2)
​

Answer 1

Answer:

300 N * 1 m + 500N *(x+1)m=500N *2m

300+500(x+1)=1000

500 (x+1)=700

x+1=7/3

x=1.4-1

x=0.4 m

Answer 2

Given: The mass of the wooden plank is 30 kg.

The two cables are separated by 2 m.

The mass of the boy is 50 kg.

To find: The distance walked by the boy.

Solution:

Let $T_{1}$ and $T_{2}$ be the tensions along the cables which are acting vertically upwards.

The weight of the wooden plank that is acting vertically downwards through the centre, 1m from either side = 300 N.

∴ $T_{1} -T_{2} =(300+500) N\\T_{1} -T_{2} = 800 N$

Let the distance walked by the boy towards the right be x m.

Moments of 300 N and 500 N forces about left end are clockwise, while that of $T_{2}$ is anticlockwise.

As the cable can sustain 500 N, $(T_{2} )_{max} = 500 N$

∴ We can say that:

$& 300\times 1+500(1+x)=500\times 2 \\ & \Rightarrow 300+500+500x=1000 \\ & \Rightarrow 800+500x=1000 \\ & \Rightarrow 500x=1000-800 \\ & \Rightarrow x=\frac{200}{500} \\ & \Rightarrow x=0.4m$

Thus, the boy can walk by to 0.4 m on either side of the centre.