Calculate the normal force for a 2 kg object at rest on an inclined surface, α=42 degrees
How much is the static coefficient of friction between the surface and the block
----
My work:
N = mg*cos(θ)
N = (2kg)(9.8n/m^2)cos(42) = 14.57 N
F = µsN
µs = F/N
µs = (2*9.8)/14.57
µs = 19.6/14.57 = 1.345
I think I got the first part of the question right, but unsure about the second part if I switched the F and N since I had to calculate the normal force at rest for the 2 kg on a flat surface first.
Thanks in advance for any help.
Oops, the units for gravity are incorrect, it should me m/s^2.
F down slope = m g sin 42
= 2 * 9.8 * sin 42
= 13.1 Newtons down the slope
so
13.1 = mu m g cos 42
or
sin 42 = mu cos 42
mu = tan 42
mu = 0.900
At least 0.900
could be more
Your calculation for the normal force is correct. The normal force (N) can be calculated as the weight (mg) multiplied by the cosine of the angle of inclination (θ). In this case, the weight is equal to the mass (2 kg) multiplied by the acceleration due to gravity (9.8 m/s^2), and the angle of inclination is 42 degrees.
N = (2 kg)(9.8 m/s^2)cos(42) ≈ 14.57 N
To calculate the static coefficient of friction (μs) between the surface and the block, you need to divide the force of friction (F) by the normal force (N).
However, in your calculations, you switched the positions of F and N. The force of friction is equal to the static coefficient of friction multiplied by the normal force (F = μsN), not the other way around.
So the correct equation should be:
μs = F/N
Since the object is at rest and not moving, the force of friction is equal to the force needed to keep the object in place, which in this case is the static friction.
Therefore, the equation becomes:
μs = static friction (F) / normal force (N)
And using the values you provided:
F = μsN
μs = F / N
μs = (2 kg)(9.8 m/s^2) / 14.57 N
μs ≈ 1.36
So the static coefficient of friction between the surface and the block is approximately 1.36.