Answer to Question #150215 in Physics for Andu

Question

The mass of the body is M = 5 kg, while the tensile force F = 50 N forms the angle a = 30 ° with the horizontal surface with coefficient of friction = 0.15 Determine the acceleration, the normal reaction force and the compressive force on the plane

Answer 1

1. Let's write down the second Newton's law in a projection to the y-axis, taking into account that the body does not move in vertical direction (acceleration $a_y = 0$ ):

$$N+F\sin30\degree - mg = 0$$

where $F = 50N$ is the tensile force, $N$ is the normal reaction force, $m = 5kg$ is the mass of the body, and $g = 9.8m/s^2$ is the gravitational acceleration. Expressing $N$, obtain:

$$N = mg - F\sin30\degree \\ N = 5\cdot 9.8- 50\cdot \sin 30\degree = 24N$$

2. According to the third Newton's law, the compressive force on the plane is equal to the normal force and equal to $24N$.

3. Let's write down the second Newton's law in a projection to the x-axis:

$$F\cos30\degree-F_{fr} = ma$$

where $F_{fr}$ is the frictional force, and $a$ is the acceleration of the body. By defitnion:

$$F_{fr} = \mu N = 0.15\cdot 24N = 3.6N$$

where $\mu =0.15$ is the coefficient of friction. Thus, obtain for acceleration:

$$a = \dfrac{F\cos30\degree-F_{fr}}{m} = \dfrac{50\cdot \cos30\degree-3.6}{5} \approx 7.9m/s^2$$

Answer. 7.9m/s^2, 24N, 24N.