A 10.0-kg uniform ladder that is 2.50 m long is placed against a smooth vertical

Question

A 10.0-kg uniform ladder that is 2.50 m long is placed against a smooth vertical wall and reaches to a height of 2.10 m, as shown in the figure. The base of the ladder rests on a rough horizontal floor whose coefficient of static friction with the ladder is 0.800. An 80.0-kg bucket of concrete is suspended from the top rung of the ladder, right next to the wall, as shown in the figure. What is the magnitude of the friction force that the floor exerts on the ladder?

The answer to the question is 538 N and I do not know how to get to it. Any help, with steps would be greatly appreciated and I will rate. Thank you.

Explanation / Answer

height = 2.1

length= 2.5 m

so theta = 57.14

fisrt let us calculate the vertical force acting downwards ,

Fy = 10*9.8 + 80*9.8 = 882 N

so , normal reaction at the bottom will be 882 N so maximum force it can withstand withoud slipping

= mu* 882= 705.6 N

now , we need to find the horizontal force exerted bt the smmoth wall ,

so we take moment at the the botoom to be zero,

so we will have,

1.25* 10*9.8cos( 57.14) - 2.5* F + 2.5*cos(57.14)*80* 9.8 = 0

=>65.87+1063.47 = 2.5* F

=>F= 451.97 N

Fh cos(90-57.14) = F

=>Fh= 451.97/ cos(32.86)=538.060 N

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