A chair of weight 150 lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of = 38.0 directed at an angle of 41.0 below the horizontal and the chair slides along the floor.Using Newton's laws, calculate , the magnitude of the normal force that the floor exerts on the chair.
You posted this twice. I answered it elsewhere. You really must provide the units of force and weight along with the numbers. I assume they are in Newtons. You also should say "degrees" after 41.0
To calculate the magnitude of the normal force that the floor exerts on the chair, we need to consider the forces acting on the chair in the vertical direction.
1. Determine the vertical component of the force you applied:
- The force you applied to the chair can be split into horizontal and vertical components using trigonometric functions.
- The vertical component, Fy, can be found using the equation Fy = F * sin(θ), where F is the magnitude of the applied force and θ is the angle below the horizontal.
- Substitute the given values: F = 38.0 N, θ = 41.0°.
2. Calculate the net force in the vertical direction:
- Since the chair is not accelerating vertically, the net force in the vertical direction must be zero according to Newton's second law (ΣFy = 0).
- The net force is the sum of the vertical components of all forces acting on the chair, which in this case includes the vertical component of the force you applied and the normal force from the floor.
- Therefore, ΣFy = Fy - Fn = 0, where Fn is the magnitude of the normal force.
3. Solve for the magnitude of the normal force:
- Rearrange the equation ΣFy = Fy - Fn = 0 to isolate the normal force: Fn = Fy.
- Substitute the calculated vertical component of the applied force, Fy, into the equation.
Let's calculate the magnitude of the normal force using the provided values:
Fy = F * sin(θ) = 38.0 N * sin(41.0°)
Fn = Fy
Now, let's plug in the values and solve for Fn.