A uniform rod 1 m long weighing 100 newtons is supported horizontally on two knife edges a and b placed at 10 cm from its ends. What will be the reaction at the supports when a 40 newton weight is suspended 10 cm from the mid point of the rod?
#Moment of Force about a point
umbral matrix
buoyant plume
65 and 75
buoyant plume
It's just all about taking moment
ancient nymph
A uniform rod 1 m long weighing 100 newtons is supported horizontally on two kni...
its about balancing torque to maintain rotational equilibrium
sum of moments about a=0
0.5(40)+0.4(100)=N_B (0.8)
similarly, sum of moments about B=0
umbral matrix
OK.
umbral matrix
its about balancing torque to maintain rotational equilibrium
I would be glad if you would explain further.
I think the 40 N should be placed 10 cm on the opposite sides of the midpoint.
Since the question did not state whether the left or right side of the midpoint.
ancient nymph
I think the 40 N should be placed 10 cm on the opposite sides of the midpoint.
its ok coz there is symmetry
ancient nymph
I would be glad if you would explain further.
so torque=r X F
u must be knowing that i suppose