How Is Friction Calculated in a Ladder Problem Involving Rotational Equilibrium?

[sona1177]

Homework Statement

A house painter stands 3.0 m above the ground on a 5.0 m long ladder that leans against the wall at a point 4.7 m above the ground. The painter weighs 680 N and the ladder weights 120 N. Assuming no friction between the house and the upper end of the ladder, find the force of friction that the driveway exerts on the bottom of the ladder.

Homework Equations

net Torque=zero
Net force=zero

The Attempt at a Solution

(680 x 3.0 x cos 70) + (120 x 2.5 x cos 70)/ (5.0 x cos 20)
=170 N

My books says 180 N is the answer, what am i doing wrong? So frustrated!

 

[anathema]

Hi there,

It looks like you have the right idea, but there are a few errors in your calculation. First, the weight of the ladder should be multiplied by the distance from its center of mass to the point where it leans against the wall (in this case, 4.7 m). Also, when calculating the torque, you should use the perpendicular distance from the force to the pivot point, which in this case is the bottom of the ladder (5.0 m). Finally, when solving for the force of friction, you should use the total torque (including both the painter's weight and the ladder's weight) divided by the distance from the bottom of the ladder to the point where it leans against the wall (4.7 m).

Here is the corrected calculation:

Net torque = (680 x 3.0 x cos 70) + (120 x 4.7 x cos 70) = 1840 Nm

Net force = 680 + 120 = 800 N

Force of friction = (1840 Nm) / (4.7 m) = 391 N

I hope this helps! Remember to always double check your units and make sure you are using the correct distances and forces in your calculations. Good luck with your studies!