Question

A rod 7.0 m long is pivoted at a point 2.0 m from the left end. A downward force of 50 N acts at the left end, and a downward force of 200 N acts at the right end. At what distance to the right of the pivot can the third force of 300 N acting upward be placed to produce rotational equilibrium? Note: Neglect the weight of the rod.
(e) 3.5 m
(d) 4.0 m
(c) 3.0 m
(b) 2.0 m
(a) 1.0 m

Answer 1

Moment 1 = 50 x 2 = 100Nm

Moment 2 = 200 x 5 = 1000Nm

distance=(moment 2-moment 1)/ Third Force

= (1000-100)Nm/300N

900Nm/300N

distance =3.0m