Question

A uniform pole 5.00 m long and weighing 100 N is to be carried at its ends by a man and his son. Where should a 250-N load be hung on the pole, such that the father will carry twice the load of his son?

Answer 1

At a distance of 1.33 m 250-N load be hung on the pole, such that the father will carry twice the load of his son.

What is torque?

Torque is the force's twisting action about the axis of rotation. Torque is the term used to describe the instant of force. It is the rotational equivalent of force.

Torque about the left end ;

$\rm F_s \times 5 m -W_l (x)-W_P(2.5)=0\\\\ 5F_S-250x-W_P(2.5)=0\\\\F_m+F_s=W_L+W_P\\\\\ F_m+F_s=350 \ N$

From  the observation,

$\rm F_m=2F_s$

After rearranging equation, we get;

$\rm 3F_S=350 N\\\\ F_s=116.67 N$

Substitute the value of the obtained force we get;

x=1.33 m

Hence, at a distance of 1.33 m ,250-N load be hung on the pole, such that the father will carry twice the load of his son.

To learn more about the torque refer to the link;

brainly.com/question/6855614

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