A large crate is at rest on a ramp at a loading dock
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The resultant static friction force is equal to 20 N to the left.
Why?
I'm assuming that you forgot to write the question of the exercise, so, I will try to complete it:
"A 50-n crate sits on a horizontal floor where the coefficient of static friction between the crate and the floor is 0.50 . A 20-n force is applied to the crate acting to the right. What is the resulting static friction force acting on the crate?"
So, if we are going to calculate the resulting static friction force, it means that there is no movement, we must remember that the friction coefficient will give us the maximum force before the crate starts to move.
We can calculate the static friction force by using the following formula:
Since the crate is not moving (static), the static friction force acting on the crate will be equal to the applied force.
Calculating we have:
Hence, the static friction force is equal to 20 N to the left (since the applied force is acting to the right)
So,
Since the static friction force is equal to the applied force, the crate does not start to move.
Have a nice day!
The intensity is defined as the ratio between the power emitted by the source and the area through which the power is calculated:
(1)
where
P is the power
A is the area
In our problem, the intensity is
. At a distance of r=6.0 m from the source, the area intercepted by the radiation (which propagates in all directions) is equal to the area of a sphere of radius r, so:
And so if we re-arrange (1) we find the power emitted by the source:
where
P is the power
A is the area
In our problem, the intensity is
And so if we re-arrange (1) we find the power emitted by the source: