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A woman pulls a suitcase..what is the angle with the horizontal? 
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#1
Oct1607, 11:35 PM

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1. The problem statement, all variables and given/known data
A woman at the airport tows her 20.0kg suitcase at constant speed by pulling on strap angle theta above the horizontal. She pulls on the strap with 35.0N force. The frictional force on suitcase is 20.0N a) what angle does the strap make with the horizontal? b) What normal force does the ground exert on the suitcase? 2. Relevant equations [tex]\sum Fx= max= Fcosf = 0[/tex] [tex]\sum Fy= may= N + Fsin theta  mg = 0[/tex] F= ma 3. The attempt at a solution a.) I know that the equations are these to get the solution of the angle however I don't know why the Fy is what the equation states it is. (the equation was given by my teacher and I didn't understand why he gave this equation since it wasn't explained) The thing is..focusing on [tex]\sum Fy= may= N + Fsin theta  mg = 0[/tex] I do know why it's = to Nmg but why is it adding Fsin theta? b.) For this part I'm kind of confused as to the way they state the question. "what normal force does the ground exert on the suitcase?" normal force is usually what the suitcase exerts on the ground which is :F= mg but wouldn't the normal force the ground exert on the suitcase be equal but opposite in direction? thus.. N suitcase= N ground m(g)= m(g) Is this fine? Thanks alot 


#2
Oct1607, 11:52 PM

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#3
Oct1607, 11:53 PM

P: 2

a.)
I know that the equations are these to get the solution of the angle however I don't know why the Fy is what the equation states it is. (the equation was given by my teacher and I didn't understand why he gave this equation since it wasn't explained) The thing is..focusing on I do know why it's = to Nmg but why is it adding Fsin theta? you add it because that is the component of the woman's pulling force in the ydirection. It acts in the same direction as the normal force. 


#4
Oct1707, 12:23 AM

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P: 824

A woman pulls a suitcase..what is the angle with the horizontal?
For part b.) so...is it the normal force that the ground is exerting on the suitcase = to
N= mg F sin theta ? I assume it would be but would the normal force be equal magnitude just oposite direction thus N? Thanks 


#5
Oct1707, 01:16 AM

P: 16

To start with, this is very similar to a problem I had in my college physics course not too long ago. This time the given information was different so it took some tweaking to get the answers and I kept getting confused. However, when I reread the problem carefully, it made a whole world of difference. I'll go ahead and explain below.
a)[tex]\Sigma F_{x} = F  f = ma_{x}[/tex] Once you understand the steps, you'll probably kick yourself for not seeing it. It just takes some algebra to get there. Given: [m]*** = 20kg [T]ension = 35N [f]riction force = 20N [a]cceleration (on the xaxis) = 0 [tex]m/s^{2}[/tex] The letter in the boxes are what I will be using to denote each quantity in the equations. Now, force F is actually the cosine of the tension on the rope, and this will be detonated as: [tex]F = Tcos\theta[/tex] We also now that the suitcase is moving at a constant speed, which means that it's acceleration is zero. Now, we can enter in the equation. [tex]Tcos\theta  f = ma_{x}[/tex] From here, we want to move f to the right to help isolate theta. [tex]Tcos\theta = ma_{x} + f[/tex] Now, we can also pull T off by divided both sides by T [tex]cos\theta = \frac{ma_{x} + f}{T}[/tex] That only leaves us with cosine theta on the left. To completely isolate theta, we have to use the inverse of cosine on the right. [tex]\theta = cos^{1}(\frac{ma_{x} + f}{T})[/tex] From here, we can enter in all of the known data we have and find theta. [tex]\theta = cos^{1}(\frac{(20kg)(0m/s^{2}) + (20N)}{35N})[/tex] [tex]\theta = 55.15^{o}[/tex] Now we can go onto part B b) Since we now have the angle theta, B becomes REALLY easy. Using your equation that your teacher gave you (which I will also explain why it is zero), you can enter in the known values to find the answer. [tex]\Sigma F_{y} = W + N + Tsin\theta = 0[/tex] This is essentially your same equation, just I have have weight represented as 'W' instead of 'mg'. Now, the reason for the equation being equal to zero is this: When you consider that the force the suitcase exerts on the ground is equal to its weight times the force of gravity, the ground must exert an equal and opposite force according to Newton's Third Law. That is where the normal force comes in or force N. In this problem, the two forces are equal, however, you must also consider that the woman is pulling the suitcase up slightly, effectively reducing the force that the suitcase is exerting on the ground and in turn, reduces the normal force. Now, I've represented this in the equation above. The weight is negative because it is pushing down while the normal force and the force of the vertical tension are both pushing up. To find the normal force, we must essentially do the same thing we did to isolate theta in part A. To do this, simply move everything over to the right side, leaving N all by its lonesome. [tex]N = W  Tsin\theta[/tex] Because both quantities moves across they change their signs, so W becomes positive and TsinTheta becomes negative. From here, we can enter in our known quantities and find N. Also remember that W = mg [tex]N = mg  Tsin\theta[/tex] [tex]N = (20kg)(9.8m/s^{2})  (35N)sin(55.15^{o})[/tex] [tex]N = 167.28N[/tex] And that wraps it up, if you have any more questions, feel free to message me. P.S. Sorry for the essaylike response 


#6
Oct1707, 01:23 AM

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P: 824

Wow..um that makes everything clearer..now if only my teacher explained it all like that..
thanks very much firefly 1 


#7
Oct1707, 01:36 AM

P: 16

No problem. Glad it makes sense to you now. If you have any more questions on further assignments, feel free to ask me.



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