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## Homework Statement

A 43.4 kg, 3.2 m uniform ladder leans against a frictionless wall.* A 88.6 kg person is standing on the ladder down 1.18 m from the top of the ladder.* The ladder makes an angle of 53 degrees with the horizontal.* What is the minimum coefficient of static friction between the ladder and the ground so that the ladder does not slip?

## Homework Equations

[tex]\sum[/tex]F

_{x}=0

[tex]\sum[/tex]F

_{y}=0

[tex]\sum[/tex][tex]\tau[/tex]=0

F

_{gx}= mgsin[tex]\vartheta[/tex]

F

_{gy}= mgcos[tex]\vartheta[/tex]

F

_{f}= [tex]\mu[/tex]*(normal force)

## The Attempt at a Solution

So the way I saw this was, if you have the person on the ladder and it begins sliding down the vertical (toward the left on the horizontal) then the force of friction would inadvertently have to be to the right. (opposing the direction of motion). To find the coffecient of static friction, I'll need the value for the force of friction and the normal force. I am unclear on how to find the force of friction but I am assuming it can be isolated out when doing the [tex]\tau[/tex] = 0 step.

## Homework Statement

A 200.00 kg uniform, horizontal beam is hinged at one end and at the other is supported by a cable that is at 13.3 degrees to the vertical.* * The beam is 2.00 m long.* Calculate the direction of the force at the hinge (measured with respect to the horizontal.

## Homework Equations

Sum of all angles in a triangle = 180.

## The Attempt at a Solution

So the vertical is perpendicular to the horizontal beam, making an angle of 90 degrees. We also know that the angle between the vertical and the supporting cable is 13.3 degrees. As such, 180-90-13.3 = 76.7.

Is this correct reasoning?

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