A crate that weighs w = 110 N is located on a rough surface. When a force Fapplied = 40 N is exerted on the crate the crate does not move. (picture is attached) (a) Keeping in mind that the crate is not moving, the net force acting on the crate is: Zero (b) According to your answer for part (a), what must be the frictional force that acts on the crate? f = 40 N (c) What is the normal force, N, acting on the crate? N =110
A crate that weighs w = 110 N is located on a rough surface. When a force Fapplied = 40 N is exerted on the crate the crate does not move. (picture is attached)
(a) Keeping in mind that the crate is not moving, the net force acting on the crate is: Zero
(b) According to your answer for part (a), what must be the frictional force that acts on the crate? f = 40 N
(c) What is the normal force, N, acting on the crate? N =110
(d) When a force Fapplied = 85 N is exerted on the crate, the crate is on the verge of moving? Under this circumstance the crate is experiencing the maximum static frictional force available. If any additional force is applied, the crate will begin to move. What is the frictional force acting on the crate under this condition? (The crate is still not moving but it is just above to move.) f maxs = 85 N
(e) (e) The maximum static static frictional force, f maxs, is related to the coefficient of friction, ?, and the normal force, N, according to the following
maximum static friction = (coefficient of friction) X (Normal force)
or more specifically
f maxs = ?s X N
We can solve for the coefficient of friction by rearranging this expression to solve for the coefficient
?s =
| f maxs |
| N |
Using your results from parts (d) [f maxs] and (c) [Normal] solve for the coefficient of static friction. Keep at least 3 digits in your answer.
? = _______
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