Question #298173

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
2022-02-16T08:34:40-0500

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

=1-Actual mechanical advantageIdeal mechanical advantage\frac{\text{Actual mechanical advantage}}{\text{Ideal mechanical advantage}}

=156.78283.9=0.8.=1-\frac{56.78}{283.9}\\=0.8.

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