The Vector Nature of Newton's Second Law

AI Thread Summary
To determine the force of friction acting on a crate being pulled with a force of 300.0 N at an angle of 30 degrees above the horizontal, only the horizontal component of the applied force is relevant. The crate moves at a constant velocity of 0.962 m/s, indicating that the net force is zero and the friction force equals the horizontal component of the pulling force. The vertical forces, including the weight of the crate and the normal force from the floor, do not affect the calculation of friction in this scenario. Therefore, the mass and velocity of the crate are not necessary for solving this specific problem. The focus remains on resolving the applied force into its components to find the friction force.
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Homework Statement


A crate of mass 50.0 kg is pulled across a level concrete floor by a force of 300.0 N acting 30.0 degrees above the horizontal. The crate moves at a constant velocity of 0.962 m/s. What is the force of friction acting on the crate?

Homework Equations


SINE, COSINE, TANGENT
F = ma
a = f/m
Fnet = F1 + F2


The Attempt at a Solution


I'm not sure how the mass and velocity comes into play in this question. Because can't I just find all the sides of the triangle to find the force of friction?
 
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\vec{F}= m\vec{a}
and the acceleration vector is 0. There are FOUR forces acting on the crate but only two of them are relevant to this question:
(1) 300N at 30 degrees above the horizontal. What are its horizontal and vertical components?
(2) Friction force which is purely horizontal. That's what you want to find.
The two forces that are not directly relevant are
(3) The weight of the crate which is purely vertical.
(4) The force of the floor on the crate which is purely vertical.

The last two are not relevant because you are asked only for a horizontal force.
Because of that, the mass and velocity do NOT play any part in this problem.
 
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