A force of 16.0 newtons is sufficient to set a box that is at rest into motion. If the coefficient of static friction is 0.380, what is the normal force acting on the box?
42.1
16 = .380 * Normal force
uuuuhhhh ... NO THE ANSWER IS NOT 16... GOT IT WROOONG!
To find the normal force acting on the box, we need to know the weight of the box and the angle at which the box is inclined. However, since the problem does not provide this information, we can assume that the box is on a horizontal surface, which means there is no angle involved.
Given that the force required to set the box into motion is 16.0 newtons, this force is equal to the maximum static friction force (the force required to keep the box at rest) since the box is initially at rest.
The formula for static friction force (F_static) is given by:
F_static = coefficient of static friction (μ_static) × normal force (N)
We can rearrange this equation to solve for the normal force (N):
N = F_static / μ_static
Plugging in the given values:
N = 16.0 newtons / 0.380
N ≈ 42.1 newtons
Therefore, the normal force acting on the box is approximately 42.1 newtons.