Respuesta :
The magnitude of this horizontal force can be calculated as :
F=mg
2x9.8=19.6N
19.6cos 30= 17 Newtons
F=mg
2x9.8=19.6N
19.6cos 30= 17 Newtons
Answer:
Horizontal force is 16.97 N.
Explanation:
It is given that,
Mass of the block, m = 2 kg
It is on an incline at a 60 degree angle is held in equilibrium by a horizontal force. We need to find the magnitude of horizontal force. The force acting on the block is its weight mg.
The horizontal component of force is, [tex]F_x=mg\ sin\theta[/tex]
The vertical component of force is, [tex]F_y=mg\ cos\theta[/tex]
So, the horizontal component of force is,[tex]F_x=2\ kg\times 9.8\ m/s^2\ sin(60)[/tex]
[tex]F_x=16.97\ N[/tex]
So, the horizontal component of force is 16.97 N.