A man pushes a chair across the floor by applying a force of 155N at an angle of 25-degrees South of East. The chair accelerates to the right along the floor at 1.20 m/s.
So far I got this
Fg = mg (m)(9.81) = ?
Fa = 155 N 25 Degrees S of E
Fax = 155cos(25) = 140.777 N
Fay = 155sin(25) = 65.50 N
a = 1.20 m/s/s
FN = ?
Fk = μFN = ?
μ = ?
Fnet = ma
How do I find FN? Is it the same as Fg? or Is it the same as Fay?
Since the floor is level, FN = Fg
The angles just indicate direction, not an incline.
Ah ok, but how do I find the mass then? Because it is not given.
F = ma
You have the force and the acceleration, so ...
To find the normal force (FN), you need to consider the forces acting on the chair in the vertical direction. The normal force is the force exerted by a surface to support an object in contact with it and acts perpendicular to the surface.
In this scenario, there are two forces acting on the chair in the vertical direction: the weight (Fg) and the vertical component of the applied force (Fay). The normal force (FN) counters the weight of the chair, preventing it from sinking into the floor.
Given that the chair is accelerating to the right, we can assume that the vertical forces are balanced. Therefore, the vertical component of the applied force (Fay) must be equal to the weight (Fg).
Since you already calculated Fay as 65.50 N, you can use this value as the normal force (FN). In this case, the normal force is not the same as Fg, but it is equal to the vertical component of the applied force.
Keep in mind that the normal force can change depending on the situation. For example, if there were an incline or additional external forces, the normal force would be different. However, in this specific scenario, the vertical component of the applied force (Fay) serves as the normal force (FN).