Question

An object sits stationary on a slpoe. The weight of the object is 56N. The component of weight at right angles to the slope is 50,3N. The component of the weight acting parallel to the slope is 24,5N. What is the angle of the slope to the horizontal?

Answer 1

The weight of the object at the inclined plane can be resolved into two components: perpendicular ($W_{\perp}=Wcos\theta$) and parallel ($W_{||}=Wsin\theta$) to the slope. Then, we can find the angle of the slope from either of this formulas:

$$\theta=sin^{-1}(\dfrac{W_{||}}{W})=sin^{-1}(\dfrac{24.5\ N}{56\ N})=26^{\circ}.$$