Question

A 100N weight rest on a 30 degrees inclined plane.Neglecting friction,how much pull must one exert to bring the weight up the plane?

Answer 1

A 100N weight rest on a 30 degrees inclined plane. Neglecting friction, how much pull must one exert to bring the weight up the plane?

**Solution:**

$$\mathrm{mg} = 100\mathrm{N} - \text{weight of the body};$$ $$\alpha = 30{}^{\circ} - \text{angle of the plane with the horizontal};$$

The first law of equilibrium along the X axis:

$$F - mg_x = 0$$ $$F = mg_x$$

From the right triangle ABC:

$$\sin \alpha = \frac{mg_x}{mg}; \quad mg_x = mg \cdot \sin \alpha$$

(2)in(1):

$$F = mg \cdot \sin \alpha = 100\mathrm{N} \cdot \sin 30{}^{\circ} = 100\mathrm{N} \cdot 0.5 = 50\mathrm{N}$$

**Answer:** to bring the weight up the plane we must act with the force bigger than 50N.