A chair of weight 85.0 N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F = 44.0 N directed at an angle of 42.0 degrees 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.

Resolve the force F=44N into the vertical component Fsin(θ), and the horizontal component Fcos(θ).

Since the angle &theta is below the horizontal, the vertical components tends to lift the chair. So the vertical reaction equals weight-vertical component.

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To find the magnitude of the normal force (n) that the floor exerts on the chair, we can use Newton's second law, which states that the net force acting on an object is equal to the product of its mass and acceleration. In this case, the chair is not accelerating vertically, so the net force in the vertical direction must be zero.

The normal force (n) is the force exerted by a surface to support the weight of an object resting on it. In this case, the weight of the chair is 85.0 N, which means the normal force is also 85.0 N to balance the chair's weight.

Now, let's break down the forces acting on the chair:

1. Weight (W): The weight of the chair is acting vertically downwards and has a magnitude of 85.0 N.

2. Applied force (F): You are pushing the chair with a force of 44.0 N at an angle of 42.0 degrees below the horizontal. To determine the effective force in the vertical direction, we need to find the vertical component of this force.

The vertical component of the applied force can be calculated as follows:
Vertical component = F * sin(θ)
= 44.0 N * sin(42.0 degrees)
= 44.0 N * 0.6691
= 29.3894 N

3. Normal force (n): The normal force acts perpendicular to the surface of contact between the chair and the floor. Since the chair is not accelerating vertically, the normal force must balance the vertical forces acting on the chair (weight and vertical component of the applied force).

Therefore,
n (magnitude of the normal force) = weight of the chair + vertical component of the applied force
= 85.0 N + 29.3894 N
= 114.3894 N

So, the magnitude of the normal force (n) that the floor exerts on the chair is approximately 114.4 N.