Question

A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of 60° with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is ______. (take g = 10 m/s²)

Options

• 100 N

• `100sqrt3 N`

• 200 N

• `200sqrt3 N`

Answer 1

A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of 60° with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is `bbunderline(100sqrt3 N)`.

Explanation:

Since the rod is in equilibrium, 

ΣF_Y = 0 

⇒ N₁ = mg = 200 N

ΣF_X = 0 

⇒ N₂ = f

ΣT_A = 0 

`Mg(1/2 cos 30°) = N_2(L cos 60°)`

`(Mg)/2 sqrt3/2 = N_2/2`

`N_2 = sqrt3/2 xx 200`

= `100sqrt3` = f