The Vector Nature of Newton's Second Law

[Turquoise]

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?

 

[HallsofIvy]

[tex]\vec{F}= m\vec{a}[/tex]
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.

 

[thomate1]

I can provide an explanation for the vector nature of Newton's Second Law in this scenario. Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. In this case, the crate is being pulled by a force of 300.0 N at an angle of 30.0 degrees above the horizontal, which means that this force can be broken down into two components: one acting in the horizontal direction and one acting in the vertical direction. The horizontal component of the force is responsible for the crate's constant velocity, while the vertical component of the force is counteracted by the normal force of the floor.

In order to find the force of friction acting on the crate, we need to consider the horizontal force component. Using the equation F = ma, we can rearrange it to solve for the force of friction (f) by substituting the mass (m) and acceleration (a) with the given values of 50.0 kg and 0.962 m/s respectively. This gives us f = ma = (50.0 kg)(0.962 m/s^2) = 48.1 N. This is the force of friction acting on the crate in the opposite direction of its motion, which is equal in magnitude to the horizontal component of the applied force.

In conclusion, the vector nature of Newton's Second Law allows us to break down a force acting on an object into its x and y components, and use this information to calculate the acceleration and other relevant quantities. In this case, we were able to determine the force of friction acting on the crate by considering the horizontal component of the applied force.