Physics Trig Problem

Hckyplayer8

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Jun 9, 2019
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I would like some guidance on how to setup this problem.

The solution for part (a) comes down to deriving the x and y components of the vector. If the woman is pulling on the suitcase with a force of 32.5N (which I believe is the hypotenuse of my right triangle) deriving just one of the x/y components allows me to use the inverse sine/cosine functions (or inverse tangent if both are found) to get the angle.

Which equation should I be using to derive the components?

1.JPG
 
[MATH]F_x=F\cos(\theta)[/MATH]
[MATH]F_y=F\sin(\theta)[/MATH]
Construct a free body diagram, and since the object is at equilibrium (how do we know this?), all opposing forces will sum to zero. What do you find?
 
[MATH]F_x=F\cos(\theta)[/MATH]
[MATH]F_y=F\sin(\theta)[/MATH]
Construct a free body diagram, and since the object is at equilibrium (how do we know this?), all opposing forces will sum to zero. What do you find?

We know the object is at equilibrium because there is no acceleration/velocity is constant.

I composed a free body diagram. An scanned upload would be best but a description will have to do for now.

The center block is the 12kg suitcase. A force vector exits and points to the left of the suitcase's travel (to the right) which is the frictional force of 20N.
A force vector extends from the suitcase in the positive vertical which an unspecified Normal Force.
A force vector extends from the suitcase in the negative vertical which is the Gravitational Force.
Lastly, I have a force vector extending diagonally from the upper right corner of the suitcase which is the Force of the lady pulling, 32.5N.

The gravitational force should just be standard gravitational acceleration, 9.80 times the mass of the object. 9.80 m/s^2 x 12kg = 117.6N.
The normal force would also be 117.6N.
 
Okay, what I would write is the following:

Sum of forces in the horizontal, with movement in the positive direction:

[MATH]\sum F_x=F\cos(\theta)-f_k=0[/MATH]
This is the horizontal component of the force exerted by the woman minus the force of kinetic friction.

Sum of the forces in the vertical, with up as the positive direction:

[MATH]\sum F_y=F\sin(\theta)+n-mg=0[/MATH]
This is the vertical component of the force exerted by the woman plus the normal force minus the weight.

What does the first equation tell us about \(\theta\)?
 
Okay, what I would write is the following:

Sum of forces in the horizontal, with movement in the positive direction:

[MATH]\sum F_x=F\cos(\theta)-f_k=0[/MATH]
This is the horizontal component of the force exerted by the woman minus the force of kinetic friction.

Plugging in known values

[MATH]\sum F_x=F\cos(\theta)-20N=0[/MATH]
which means

[MATH]\sum F_x=F\cos(\theta)=20[/MATH]
Sum of the forces in the vertical, with up as the positive direction:

[MATH]\sum F_y=F\sin(\theta)+n-mg=0[/MATH]
This is the vertical component of the force exerted by the woman plus the normal force minus the weight.

What does the first equation tell us about \(\theta\)?

Again, plugging in the known values.

[MATH]\sum F_y=F\sin(\theta)+117.6kg-(9.80m/s^2)(12kg)=0[/MATH]
I do not know what the first equation tells us about theta.
 
I would write:

[MATH]\theta=\arccos\left(\frac{f_k}{F}\right)[/MATH]
We are given the values on the right, so we know we can compute \(\theta\). Now, this tells us:

[MATH]\sin(\theta)=\frac{\sqrt{F^2-f_k^2}}{F}[/MATH]
And so the normal force is:

[MATH]n=mg-\sqrt{F^2-f_k^2}[/MATH]
Again, we know all the values on the right. Does this make sense?
 
I would write:

[MATH]\theta=\arccos\left(\frac{f_k}{F}\right)[/MATH]
We are given the values on the right, so we know we can compute \(\theta\). Now, this tells us:

[MATH]\sin(\theta)=\frac{\sqrt{F^2-f_k^2}}{F}[/MATH]
And so the normal force is:

[MATH]n=mg-\sqrt{F^2-f_k^2}[/MATH]
Again, we know all the values on the right. Does this make sense?

I got the answer but I'm not sure why the above would come intuitively to me. Based on the given diagram, it was obvious what the hypotenuse of the triangle was. But why would I know the frictional effect vector was the adjacent? I thought the frictional vector was an effect on the suitcase, not the lady.

Then I'm lost on what (I presume are) trig identities used to derive the second equation.
 
In the horizontal direction there are two opposing forces, the horizontal component of the force exerted by the woman, and the frictional force. In the vertical direction there are two forces pointing up, the normal force and the vertical component of the force exerted by the woman, and there is one force pointing down, which is the weight of the suitcase. Do you have these in your free body diagram?
 
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