A chair of weight 75.0N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F = 45.0N directed at an angle of 40.0¡ã below the horizontal and the chair slides along the floor.
Using Newton's laws, calculate n, the magnitude of the normal force that the floor exerts on the chair.
Express your answer in newtons.
To calculate the magnitude of the normal force (n) exerted by the floor on the chair, we need to consider the forces acting on the chair in the vertical direction.
The weight of the chair, acting vertically downward, is given as 75.0 N.
The force you applied to the chair is directed 40.0 degrees below the horizontal and has a magnitude of 45.0 N. Let's break this force into its vertical and horizontal components.
Vertical component: Fv = F * sin(angle)
Fv = 45.0 N * sin(40.0 degrees)
Fv ≈ 45.0 N * 0.643
Fv ≈ 28.935 N
Horizontal component: Fh = F * cos(angle)
Fh = 45.0 N * cos(40.0 degrees)
Fh ≈ 45.0 N * 0.766
Fh ≈ 34.47 N
Now, we can apply Newton's second law in the vertical direction:
Sum of vertical forces = ma, where a = 0 (since the chair does not accelerate vertically)
The sum of vertical forces includes the weight of the chair (mg) and the vertical component of the force you applied to the chair (Fv), both acting in opposite directions:
Sum of vertical forces = n - mg - Fv = 0
We can rearrange this equation to solve for n:
n = mg + Fv
Substituting the known values:
n = (75.0 N) + (28.935 N)
n ≈ 103.935 N
Therefore, the magnitude of the normal force exerted by the floor on the chair is approximately 103.935 N.
To calculate the magnitude of the normal force (n) that the floor exerts on the chair, we can start by resolving the force applied into its horizontal and vertical components.
The force applied (F) is 45.0N directed at an angle of 40.0 degrees below the horizontal. The vertical component of the force can be found using trigonometry:
Vertical component of F = F * sin(θ)
= 45.0N * sin(40.0 degrees)
≈ 28.98N
Since there is no vertical acceleration (the chair doesn't move up or down), the vertical components of the forces must cancel each other out. The normal force (n) exerted by the floor counteracts the vertical component of the applied force, preventing the chair from sinking into the floor:
n - 28.98N = 0
Solving for n, we find:
n = 28.98N
So, the magnitude of the normal force exerted by the floor on the chair is approximately 28.98 newtons.
first, you are adding to the normal force pushing downward.
normal force=45SinTheta+mg