Question about finding the force using vector projections

In summary: So while they are not equivalent in magnitude and direction, they are related in terms of the forces acting on the cart on the ramp.In summary, the problem in the pre-calculus textbook involves finding the force required to keep a 200-pound cart from rolling down a ramp inclined at 30 degrees. This can be done by projecting the gravitational force vector onto a unit vector in the direction of the ramp. The resulting magnitude is 100 pounds. The unit of force can be expressed in different units, such as Newtons or pounds. Additionally, the force can also be calculated using trigonometric functions, with the force acting down the ramp being equivalent to the force perpendicular to the ramp. However, these two forces are not equivalent in magnitude and
  • #1
tokki1510
2
0
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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  • #2
tokki1510 said:
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.
 
  • #3
Mark44 said:
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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  • #4
tokki1510 said:
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.
 

Related to Question about finding the force using vector projections

1. What is vector projection and how is it used to find force?

Vector projection is a mathematical method used to find the component of a vector in a specific direction. In the context of force, it can be used to determine the force acting in a particular direction by projecting the force vector onto that direction.

2. Can vector projection be used to find the magnitude and direction of force?

Yes, vector projection can be used to find both the magnitude and direction of force. The magnitude of the projected vector represents the magnitude of the force acting in that direction, while the direction of the projected vector represents the direction of the force.

3. What is the formula for finding force using vector projection?

The formula for finding force using vector projection is: F = |F|cosθ, where F is the force vector, |F| is the magnitude of the force, and θ is the angle between the force vector and the direction in which the force is being projected.

4. Can vector projection be used for finding the force in three-dimensional space?

Yes, vector projection can be used to find the force in three-dimensional space. In this case, the force vector and the direction of projection will both have three components, and the formula for finding force using vector projection will be slightly different.

5. How is vector projection related to dot product?

Vector projection is closely related to dot product, as the dot product of two vectors is equal to the product of their magnitudes and the cosine of the angle between them. In fact, the formula for finding force using vector projection is derived from the dot product formula.

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