An air-conditioning unit is fastened to a roof that slopes at an angle of
35? above the horizontal . Its weight is a force on the air
conditioner that is directed vertically downward. In order that the
unit not crush the roof tiles, the component of the unit?s weight perpendicular to the roof cannot exceed 425 N. (One newton, or 1 N, is
the SI unit of force. It is equal to 0.2248 lb.) (a) What is the maximum allowed weight of the unit? (b) If the fasteners fail, the unit
slides 1.50 m along the roof before it comes to a halt against a ledge.
How much work does the weight force do on the unit during its slide
if the unit has the weight calculated in part (a)? As we described in
Section 1.10, the work done by a force on an object that undergoes
a displacement s is W=F.s

Dear Student

From the figure, the magnitude of the weight of the unit acting in the y-direction, wy is equal to the normal force.    That is, N = w cosθ  Therefore, w = N/cosθw =425cos 35 =4250.82 = 518.3 Nb)Fx = F sin θ = 518.3  sin 35 = 518.3 ×0.574 = 297.5NW = Fxs = 297.5 N×1.50m  =446.25 Nm

Regards

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