- #1

astoll

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- Homework Statement
- A 40-kg uniform ladder that is 5.0 m long is placed against a smooth wall at a height of h = 4.0 m,

as shown in the figure. The base of the ladder rests on a rough horizontal surface whose coefficient

of static friction with the ladder is 0.70. An 80-kg bucket is suspended from the top rung of the

ladder, just at the wall. What is the magnitude of the force that the ladder exerts on the wall?

- Relevant Equations
- sum of forces in the y direction = Fn - Fbucket - Fearth = 0

Sum of the forces in the X direction = Ff +Fladder -Fn = 0

Ff = usFn

Torque= Flsin(theta)

I've been working on this problem for a couple days now and I'm clearly missing something.

I first went ahead and solved the triangle. Hypotenuse is 5, height is 4, the last side is 3 and the angle is 53 degrees.

I went ahead and did the sum of forces in the y direction = Fn - Fbucket - Fearth = 0 to solve for Fn in the y direction and got (80)(9.8) + (40)(9.8) = Fn = 1176 N.

I then solved for force of friction given Ff = usFN or (0.7)(1176) = 823.2 N

I set the axis of rotation to where the bottom of the ladder meets the ground and set up torque equations.

The bottom of the ladder at the axis of rotation has no torque (tfriction, tFny =0).

Torque of the ladder is given at the center of mass of the ladder (392)(2.5)(sin37) = -589.78 N*m (clockwise)

Torque of the bucket is given by (784)(5)(sin37) = -2359.11 N*m (clockwise)

Torque of the normal Force in the x direction pushing off the wall is given by (FNx)(5)(sin53) = 3.99FN (counterclockwise)

Torque of the force of the ladder on the wall is (Fladder)(5)(sin127) = -3.99Fl (clockwise)

This gives me -2948.99 +3.99Fnx = 3.99Fl

-739 + FN = Fl

FN = - Fl

This will give me half of the answer I am looking for. Is my issue just that while there are equal and opposite forces pushing from the wall on the ladder and the ladder on the wall, they are not equal in torque? As in, the wall on the ladder will have a torque, but the ladder on the wall will not?

The correct answer is supposed to be 740 N. Any help on where I am going wrong would be appreciated.

I first went ahead and solved the triangle. Hypotenuse is 5, height is 4, the last side is 3 and the angle is 53 degrees.

I went ahead and did the sum of forces in the y direction = Fn - Fbucket - Fearth = 0 to solve for Fn in the y direction and got (80)(9.8) + (40)(9.8) = Fn = 1176 N.

I then solved for force of friction given Ff = usFN or (0.7)(1176) = 823.2 N

I set the axis of rotation to where the bottom of the ladder meets the ground and set up torque equations.

The bottom of the ladder at the axis of rotation has no torque (tfriction, tFny =0).

Torque of the ladder is given at the center of mass of the ladder (392)(2.5)(sin37) = -589.78 N*m (clockwise)

Torque of the bucket is given by (784)(5)(sin37) = -2359.11 N*m (clockwise)

Torque of the normal Force in the x direction pushing off the wall is given by (FNx)(5)(sin53) = 3.99FN (counterclockwise)

Torque of the force of the ladder on the wall is (Fladder)(5)(sin127) = -3.99Fl (clockwise)

This gives me -2948.99 +3.99Fnx = 3.99Fl

-739 + FN = Fl

FN = - Fl

This will give me half of the answer I am looking for. Is my issue just that while there are equal and opposite forces pushing from the wall on the ladder and the ladder on the wall, they are not equal in torque? As in, the wall on the ladder will have a torque, but the ladder on the wall will not?

The correct answer is supposed to be 740 N. Any help on where I am going wrong would be appreciated.

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