Consider the ladder in the figure below and imagine two scenarios. In both scena

Question

Consider the ladder in the figure below and imagine two scenarios. In both scenarios the ladder is in rotational equilibrium. (1) There is no friction between the ladder and the floor, and there is a frictional force FW between the wall and the ladder. (2) There is no friction between the ladder and the wall, and there is a frictional force FF between the floor and the ladder. Which force is larger, FW or FF? Hint: Consider the value of the angle ? in the figure below.

In this problem why is normal force from the wall not counted for torque for scenario 1?

Explanation / Answer

a)

case 1:

from the digram,

balancing the forces along the vertcal line,

NF+fw=m1*g+mp*g

NF+fw=(m1+mp)*g

fw=(m1+m2)*g-NF ------(1)

case 2:

balancing the forces along the horizontal line,

fF=Nw -------(2)

here,

fF is larger than the fw

because, effective weight (m1*g+mp*g) is downward,

hence the Nf larger and fF is larger.

b)

torque due to Normal force from the wall not counted,

because ladder is not rotating which is actually fixed

and

Nw is acts at the axis of rotation at wall and ladder.

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