To solve this problem, we will use the following equations:
1. ΣF = ma (Newton's second law)
2. F = mg (force due to gravity)
3. ΣFx = max (net force in the x-direction)
4. ΣFy = may (net force in the y-direction)
Given:
- Mass of the block (m) = 2.0 kg
- Angle of the incline (θ) = 60.0°
- There is no friction
a. Determining the magnitude of the horizontal force:
Since the block is held in equilibrium, the net force in both the x and y direction must be zero. We can start by finding the components of the weight force:
F = mg = (2.0 kg)(9.8 m/s^2) = 19.6 N (up the incline)
The weight force has two components:
- Fx = F * sin(θ) = 19.6 N * sin(60.0°) = 16.96 N (opposite to the force holding the block in equilibrium)
- Fy = F * cos(θ) = 19.6 N * cos(60.0°) = 9.80 N (perpendicular to the incline)
The horizontal force that holds the block in equilibrium is equal in magnitude and opposite in direction to Fx:
Horizontal force = -Fx = -16.96 N
b. Determining the magnitude of the normal force:
Since the block is held in equilibrium, the force normal to the incline (normal force, N) is equal in magnitude and opposite in direction to the y-component of the weight force (Fy):
Normal force = Fy = 9.80 N
Therefore, the magnitude of the normal force on the block is 9.80 N.
A 2.0 kg block on an incline at a 60.0° angle is held in equilibrium by a horizontal force.
a. Determine the magnitude of this horizontal force. (Disregard friction.)
b. Determine the magnitude of the normal force on the block.
1 answer