A uniform rod of mass 100g has a length of 1m.it is supported horizantally on two knife edges placed 10cm from it end. What will be the reaction at these supports when a 50g mass is suspended 30cm from one of the knife edges?

Answer 1

from one knife edge
center of rod is .4 meters
added mass is .3 meters
other knife edge is .8 m away from first

total mass = .1 + .05 = .15 kg

moment around close knife edge = g(.3*.05+.4*.1) = .055 g
so
F(.8) = .055 g
F = .06875 g
if g = 9.81 m/s^2
F = .674 Newtons

total weight = .15(9.81) = 1.47 N
so other F = 1.47 - .674 = .798 Newtons

Answer 2

Answers