Question

Two like parallel forces of 300N and 200N are acting at the ends of the rod of 4m length. Distance of resultant is​

Answer 1

The resultant acts at a distance of 1.6 m from 300 N force.

Explanation:

Given:

Two parallel forces 300 N and 200 N are acting at the ends of the rod of 4 m length

To find out:

The distance of the resultant

Solution:

If equilibrium condition the resultant will act at the point where moment of forces will be equal to zero.

Consider the point at which resultant of the forces will act, is 'x' m from the force 300 N.

Thus the point is (4 - x) m far from the force 200 N.

Using the formula

Torque = Distance x Force acting

$300\times x=200\times (4-x)$

$\implies 300x=800-200x$

$\implies 500x=800$

$\implies x=\frac{8}{5}$

$\implies x=1.6$ m

Therefore the resultant acts at a distance of 1.6 m from 300 N force.

Hope this answer is helpful.

Know More:

Q: Two like parallel forces are acting at a distance of 24 mm apart and their resultant is 20 N. It the line of action of the resultant is 6 mm from any given force, the two forces are:

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