QUESTION: The coefficients of static and kinetic frictions for plastic on a surf
ID: 1259747 • Letter: Q
Question
QUESTION:
The coefficients of static and kinetic frictions for plastic on a surface are 0.60 and 0.5, respectively. How much horizontal force would you need to apply to a 3.0 N plastic calculator to move it at constant velocity?
I am having some difficulty with this question. The constant velocity is throwing me off. I have dont some work, which I will show below the question, to try and help pave the way but the answer is not clicking.
WORK SO FAR:
F=ma
mass of calculator conversion= 3.0 N (101.97kg/1N)(1kg/1000g) = 0.30591 kg
Force due to gravity = 0.30591 kg * 9.8 m/s^2 = 2.998 N
Fstatic/Fnormal = mu s = 0.6
Fkinetic/Fnormal = mu k = 0.5
Fstatic = 0.6 * 2.998 N = 1.7988 N
F kinetic = 0.5 * 2.998 N = 1.499 N
If velocity will eventually be constant, I know that the acceleration will be 0. However, I cannot figure out how to solve for the horizontal force. Any help would be greatly appreciated!
Explanation / Answer
If the plastic calculator moves with constant velocity, the net force exerted on it must be zero, and the friction force always opposes the relative motion of the object. The net force along the x-axis is,
Fnet = F - Ffri
0 = F - Ffri
Hence, the horizontal force is,
F= Ffri
Similarly, the net force exerted on the plastic along the y-axis is,
Fnet,y = N - mg
0 = N - mg
Hence, the normal force is,
N = mg
The friction force when the plastic moves from rest position is,
Ffri = usN = usmg
Therefore, the horizontal force is,
F = usmg = 0.60*3.0 = 1.8 N