Solving Tension at an Angle: Friction, Force, Coefficient

[Yshai24]

Homework Statement

A student pulls her 22-kg suitcase through the airport at constant velocity. The pull strap makes an angle of 50 degrees above the horizontal. (a) If the frictional force between suitcase and floor is 75N, what force is the student exerting? (b)What's the coefficient of friction?

Homework Equations

F=ma
μk x N

The Attempt at a Solution

So, I have the answer to this in the back of the book, but I can't figure it out. Here is my attempt:

First I split tension in two components and tried to find the x and y Fnet,

y: -mg + N + T x sin50 = ma = 0 (constant velocity = zero acceleration)
x: T x cos50 - 75N = ma = 0

Algebraically reworking this:

(T x Cos50) / μk = N

so substituting this in, my final equation should be:

T=(22kg)(9.8m/s2)/(cos50/μk)+sin50

The part where I am confused is how to find the coefficient. I tried all the ways I knew and I know the hint is that the friction is 75N. I tried μk=75N/215.6N (m x g, 22kg x 9.8) = .34 but this is not correct as it isn't right in the back of the book and I know that the normal force is less because of the angle of the tension.

Ultimately, I am confused because it seems circular. I need the normal force to find the coefficient but I need the coefficient to find the normal force. Every other equation I have seen like this provides the coefficient.

Any help is greatly appreciated.

 

[Mr. Box]

I don't know how you got your formula for T, but I'm assuming you subbed in your formula for N into your y component formula. Rearranging you get T=[(22-kg)(9.8-m/s^2)-(T*Cos 50/uk)]/sin 50 , which is different from your equation for T. You can quickly tell your equation for T is wrong because you have a plus sin 50 on your RS, where in the y component equation you have a times sin 50 on the LS.

I think it would be best just to solve for T in your x component equation, then sub that into your y component equation, where you can solve N and from there solve uk. Don't use my equation for T to do this question since you have the variable T on both sides.

 

[Yshai24]

Thank you for the reply.

I am still a bit confused though. I used the equation that my teacher worked out in class. When I do that equation with the coefficient of static friction that the answer gives, my solution is correct. So I know that it is the right formula, or at least will yield the right answer. The book gives and answer of μk=.594 and Force T = 116.7N

You can see that if you use my equation, with the given coefficient, it will come out to 116.7.

So I tried to do what you said and solve for T in the x direction but I honestly don't know what T would be. Isn't that the problem, that I don't know it?

I initially thought that Tx had to be 75N because the friction is 75N. Is that correct? In that case, How would I solve for X?

Tcos50 - 75N = ma
75N x cos50 - 75N?

Still a bit lost here.

Thanks for your time.

 

[Mr. Box]

You set the x component equation to zero yourself. Thus T=75/cos 50. What I meant was in the equation labelled with "x:" you can solve T by rearranging. The x component of T is 75 N, but T with its x and y components added together is 116.7.

I see that your equation is right, but the way you wrote it is wrong:
It should be T=(22kg)(9.8m/s2)/[(cos50/μk)+sin50] , otherwise it does not look like you are dividing by sin 50, rather it looks like you are adding it to the fraction.

 

[Yshai24]

Solved

Got it. I think I was over complicating it because I started with the last equation. But it seems easier than I initially thought. Perhaps because I still am a little rusty on trig functions.

So:

Tcos50 -75n =ma = 0 is 75N/cos50 = T = 116.7N

This gives me the force that is being exerted by the student on the suitcase. Using this I can solve for Normal force of Y

-mg + N + T x sin50 = ma = 0

N+(T x sin50) = mg

N=mg - (T x sin50) = (22kg)(9.8m/s2) - (116.7 x sin50) = 126.2

So using this, I get:

μk=Fk / N = 75/126.2 = .594

Sound good?

Thanks again for the help. I need to study sin cos tan more.