A uniform ladder 6.58 m long weighing 460 N rests with one end on the ground and

Question

A uniform ladder 6.58 m long weighing 460 N rests with one end on the ground and the other end against a perfectly smooth vertical wall. The ladder rises at 57.0 degrees above the horizontal floor. A 720 N painter finds that she can climb 2.85 m up the ladder, measured along its length, before it begins to slip.

What force does the wall exert on the ladder?
Find the friction force that the floor exerts on the ladder.
Find the normal force that the floor exerts on the ladder.

ALL please, and please show work...thanks!

Explanation / Answer

Sum of forces in the y:

N2-mg-Mg = 0

N2 = mg+Mg = 460+720 = 1180N

So this is the normal force form the floor that is exerted on the ladder.

Sum of forces in y:

ff-N1=0

ff is the frictional force at the bottom; N1 is the normal force from the top of the ladder.

Take the sume of the forces about the bottom of the ladder (where ff and N2 interact):

N1*6.58sin(57)-(mg)(2.85cos(57))-(Mg)(6.58/2*cos(57)=0

N1 = [(720)(2.85cos(57))+(460)(3.29*cos(57))]/(6.58sin(57))=351.88N

ff = N1 = 351.88N

So the answers are:

351.88

351.88

1180

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