Question

a boy and a man carry a uniform rod of length l horizontally in such a way that the boy get 1/4of load.if the boy is one end of the rod,the distance of the man from the other end is​

Answer 1

Answer:l/3

Explanation:

As we need to determine the position of the man as it carries the rod with the boy, we must first determine the conditions that will let us determine that distance.

Setting the pivot point of the rod at its center of gravity, the net torque working on the system will be of zero as the rod remains in the horizontal position. We build up the following equation:

Answer 2

man stands at $x=\frac{l}{3}$ from the other end of the rod.

Explanation:

Given:

end support of load by boy, $F_B=\frac{w}{4}$

Where w = mass of the uniform rod

Then the load supported by the man at x distance from the other end, $F_M=\frac{3w}{4}$

Now for equilibrium moment about any point is zero:

$\sum M=0$

Take moment about the point of support of boy,

$\frac{3w}{4} \times (l-x)=w\times \frac{l}{2}$

Since the mass of the beam is a uniformly distributed load so while taking its moment the complete load can be considered to act at its center of gravity, i.e. 0.5l

$x=\frac{l}{3}$

TOPICS: moment of a balanced beam,

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