B Question about finding the force using vector projections

In my pre-calculus textbook, the problem states:

A 200-pound cart sits on a ramp inclined at 30 degrees. What force is required to keep the cart from rolling down the ramp?

The gravitational force can be represented by the vector F=0i-200j

In order to find the force we need to project vector F onto a unit vector in the direction of the ramp, so we use

v = cos(30°)i+(sin30°)j

After calculating the projection of F onto v, we get -100(cos(30°),(sin30°)). Taking the magnitude of this, we get that it takes a force of 100 pounds to keep the cart from rolling down the hill.

So I have a couple of questions about this.

First, why can we express force in pounds? Someone told me we can only express force in Newtons.

Second, I've seen others solve it this way:

The force required to keep the cart from rolling is 200×sin(30°)
And when they solve for the force perpendicular to the hill it is 200×cos(30°)

Why is it possible to calculate it this way as well?
 
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In my pre-calculus textbook, the problem states:

A 200-pound cart sits on a ramp inclined at 30 degrees. What force is required to keep the cart from rolling down the ramp?

The gravitational force can be represented by the vector F=0i-200j

In order to find the force we need to project vector F onto a unit vector in the direction of the ramp, so we use

v = cos(30°)i+(sin30°)j

After calculating the projection of F onto v, we get -100(cos(30°),(sin30°)). Taking the magnitude of this, we get that it takes a force of 100 pounds to keep the cart from rolling down the hill.

So I have a couple of questions about this.

First, why can we express force in pounds? Someone told me we can only express force in Newtons.
In the MKS metric system, the unit of force is the Newton. In the cgs Metric system, the unit of force is the dyne. The unit of force in the English system is the pound, so a force can be expressed in Newtons, dynes, or pounds, depending on the system of units in use in the problem.
tokki1510 said:
Second, I've seen others solve it this way:

The force required to keep the cart from rolling is 200×sin(30°)
And when they solve for the force perpendicular to the hill it is 200×cos(30°)

Why is it possible to calculate it this way as well?
The force acting down along the ramp is perpendicular to the force acting straight into the ramp. The two force vectors form the legs of a right triangle. If you draw a diagram of the forces, you should see this, and how the 30° enters into the calculation.
 
The force acting down along the ramp is perpendicular to the force acting straight into the ramp. The two force vectors form the legs of a right triangle. If you draw a diagram of the forces, you should see this, and how the 30° enters into the calculation.
In the first diagram, I used this to set up for finding the magnitude of w1 by taking the projection of F onto v.

Between the two diagrams, is w1 and AC equivalent? I thought that vectors need to have the same magnitude and direction to be equivalent.
Have I set this up correctly?

Diagram.jpg
 

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Between the two diagrams, is w1 and AC equivalent?
Both vectors have the same magnitude, but they point in opposite directions. w1 is the force of the cart acting down the ramp, and AC is the force that's needed to compensate for w1.
 

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