Answer to Question #168853 in Mechanics | Relativity for April

Question

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

Answer 1

Answer

Ideal mechanical advantage

$F_i=mgcos\theta\\=30\times9.8\times cos5°\\=283.9N$

Actual mechanical advantage

$F_a=\mu mgcos\theta\\=0.20\times30\times9.8\times cos5°\\=56.78N$

So efficiency

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

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