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.
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.