Magnitude of normal force question

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SUMMARY

The magnitude of the normal force acting on the suitcase is calculated to be 288N using the equation N = mg - F_vertical, where F_vertical is the vertical component of the applied force (105N at 38 degrees). The coefficients of static and kinetic friction are 0.273 and 0.117, respectively. To determine if the suitcase will slide, the applied horizontal force (105N) must exceed the force of kinetic friction (33.696N), which it does, confirming that the suitcase will move. The discussion emphasizes the importance of understanding the difference between static and kinetic friction in motion scenarios.

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JadeLove
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Homework Statement



A woman pulls on her suitcase with a force of 105N at an angle of 38 degrees above the horizontal. The coefficients of static and kinetic friction between and suitcase and the floor are 0.273 and 0.117. The mass of the suitcase is 36 kg. what is the magnitude of the normal force acting on the suitcase due to the floor?

Homework Equations



normal force - weight + vertical component =0

The Attempt at a Solution



N - mg +105 sin 38=0

N=36kg*9.8m/s/s - 105 sin 38 = 288N

So I have figured out the magnitude of the normal force, but I don't know if the suitcase will slide. How do I figure that out?
 
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Well for the suitcase to move, her horizontal force must be greater than the force of friction. How does friction relate to the normal force and which coefficient of friction would determine if it will move, static or kinetic?
 
rock.freak667 said:
Well for the suitcase to move, her horizontal force must be greater than the force of friction. How does friction relate to the normal force and which coefficient of friction would determine if it will move, static or kinetic?

Okay, but I'm given two different frictions, but I chose to work with kinetic because I know it's to do with motion... Or at least pretty sure.

Ff = u(288N)
Ff = 0.117 x 288N
Ff = 33.696

That's so little :S So since she's pushing with 105 N, the suitcase DOES move, correct??
 
JadeLove said:
Okay, but I'm given two different frictions, but I chose to work with kinetic because I know it's to do with motion... Or at least pretty sure.

Ff = u(288N)
Ff = 0.117 x 288N
Ff = 33.696

That's so little :S So since she's pushing with 105 N, the suitcase DOES move, correct??

The kinetic friction is used when the body is already moving, in order for motion to occur you'd need to overcome static friction first. But once Fapplied<Ff, then no motion will occur.
 
Thank you makes sense, I got my answer my brother helped & showed me. Although I think you could have been a liiiiittle more like, well :D the formula for calculating Fs is etc, you know? Because you're too smart, & I don't really get it physics language, but now I get what you meant form the starts, thanks ^___^

Jade
 
JadeLove said:
Thank you makes sense, I got my answer my brother helped & showed me. Although I think you could have been a liiiiittle more like, well :D the formula for calculating Fs is etc, you know? Because you're too smart, & I don't really get it physics language, but now I get what you meant form the starts, thanks ^___^

Jade

But you knew what to do! You started correctly by calculating the normal force utilizing the vertical component. I tried to make it simple by how I wrote the second post. (you had all the formulas correct, so I assumed you knew had all the formulas correct)
 

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