Question

Find the magnitude of two like parallel forces acting at a distance of 240 mm, whose resultant is200 N, and its line of action is at a distance of 60mm from one of the forces​

Answer 1

Answer:

Explanation:

Let F1 and F2 are two parallel force acting at a distance of 240 mm apart and their resultant is 200N.

so, √{F1² + F2² +2F1.F2cos0°} = 200N [ as forces are parallel so, angle between them is 0°

or, √(F1 + F2)² = 200N

or, F1 + F2 = 200N----(1)

now, the line of action of the resultant is 6mm from the given force.

if we assume F1 > F2 , then line of action will be near of F1 . Let R is the point where line of action is appear.

i.e., F1-(06mm)-R-(180mm)-F2

net torque at R should be zero.

so, 0 = F1 × 60mm(→) + F2 × 180mm(←)

or, 60F1 - 180F2 = 0----(2)

from equations (1) and (2),

F1 = 150N and F2 = 50N.