Question #168853

An inclined plane as shown in figure 1 is used to raise an object of mass 30 kg. If the plane is

inclined 5° above the horizontal and the coefficient of friction is 0.20, calculate the ideal

mechanical advantage, actual mechanical advantage, and the efficiency of this machine


1
Expert's answer
2021-03-09T07:44:42-0500

Answer

Ideal mechanical advantage

Fi=mgcosθ=30×9.8×cos5°=283.9NF_i=mgcos\theta\\=30\times9.8\times cos5°\\=283.9N

Actual mechanical advantage

Fa=μmgcosθ=0.20×30×9.8×cos5°=56.78NF_a=\mu mgcos\theta\\=0.20\times30\times9.8\times cos5°\\=56.78N

So efficiency

=Actual mechanical advantageIdeal mechanical advantage\frac{\text{Actual mechanical advantage}}{\text{Ideal mechanical advantage}}

=56.78283.9=0.2=\frac{56.78}{283.9}\\=0.2



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