A750 n painter needs to climb 2.75 m up a ladder (measured along its length), without the ladder slipping. the uniform ladder is 7.0 m long and weighs 450 n. it rests with one end on the ground and the other end against a perfectly smooth vertical wall. the ladder rises at an angle of 60.0 degrees above the horizontal floor. what is the friction force and normal force that the floor must exert on the ladder? known: unknown: strategy solution:

Explanation:

Total vertically down ward force =

750 + 450

= 1200 N

Total force acting vertically upwards that is reaction force from the ground on the leg of the ladder is R

Since there is no vertical movement , these two forces will be equal ie

Reaction force from the ground R = 1200 N

Similarly there are two forces acting in horizontal direction

1 ) Frictional force acting inwards towards the wall

2 ) The reaction force from the wall on the ladder acting away from the wall on the point where ladder touches the wall.

Since there is no horizontal movement in the ladder therefore these two forces too will be equal in magnitude and acting in opposite direction. Let this force be F.

Now we shall take moment of forces acting on the ladder about the point where ladder touches the ground.

Torque due to weight of the ladder

= 450 x 7 /2 cos 60 = 787.5 N m

Torque due to weight of the painter

750 cos 60 x 2.75 = 1031.25 Nm

Torque due to friction will be zero.

torque due to reaction force of the vertical wall

= F X 7 Sin 60

For equilibrium

450 x 7 /2 cos 60 + 750 cos 60 x 2.75 =  F X 7 Sin 60

787.5 +1031.25 = F X 7 Sin 60

F = 300 N

So friction force = 300 N .