Friction and torque of ladder resting on wall

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SUMMARY

The discussion focuses on the calculation of torque for a uniform ladder resting against a smooth wall, with specific emphasis on the forces acting on the ladder, including the weight of the ladder and the do-it-yourself enthusiast. The key equation derived is -m1g(L-d)cos(x)-(m2g)(L/2)(cos(x))-fLsin(x)+N2Lcos(x), which incorporates the mass of the ladder (m2), the mass of the enthusiast (m1), and the angle (x) with the ground. The participants clarify that torque is determined by the force multiplied by the perpendicular distance to the pivot point, explaining the relevance of vertical and horizontal distances in this context.

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  • Concept of normal force in physics
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SA32
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Homework Statement


A uniform ladder with mass m2 and length L rests against a smooth wall. A do-it-yourself enthusiast of mass m1 stands on the ladder a distance d from the bottom (measured along the ladder). The ladder makes an angle x with the ground. There is no friction between the wall and the ladder, but there is a frictional force of magnitude f between the floor and the ladder. N1 is the magnitude of the normal force exerted by the wall on the ladder, and N2 is the magnitude of the normal force exerted by the ground on the ladder. Throughout the problem, consider counterclockwise torques to be positive. None of your answers should involve pi (i.e., simplify your trig functions).

Sum the torques given in the problem about the point at which the ladder touches the wall.

Homework Equations


T(torque)=F*d

See attachment for diagram.

The Attempt at a Solution


I know the answer is,

-m1g(L-d)cos(x)-(m2g)(L/2)(cos(x))-fLsin(x)+N2Lcos(x)

I know that torque is the force multiplied by the distance to the pivot point. What I don't get is why we are only concerned with with each force's vertical distance from the pivot point and not the horizontal distance. Help, please?
 

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SA32 said:
I know that torque is the force multiplied by the distance to the pivot point.
Actually, torque is the force multiplied by the perpendicular distance to the pivot point--direction matters!
What I don't get is why we are only concerned with with each force's vertical distance from the pivot point and not the horizontal distance.
Since it's the perpendicular distance that counts: for vertical forces (like weight and normal force), it's the horizontal distance; but for horizontal forces (like friction), it's the vertical distance.
 
Ahh, that makes sense. Thanks!
 

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