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Tenelen
#1 Sep 15 2009 05:25pm
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I've got my answer, but I want to make sure I'm doing it right, so somebody else do it.

At the instant a race began, a 65-kg sprinter exerted a force of 720N on the starting block at a 22 degree angle with respect to the ground. (a) What was the horizontal acceleration of the sprinter? (b) If the force was exerted for 0.32s with what speed did the sprinter leave the starting block?


All attempts welcome, just show how you came to the conclusion so I can see what's going on please.
ToujoursLaVerite
#2 Sep 15 2009 05:34pm
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Check your PM box.
Terminsel
#3 Sep 15 2009 05:35pm
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(a)
F = m * a
m = 65kg
F = 720
a = 11.07 m/s^2
Since its at a 22 degree angle we want the cosine of 22 * 11.07 to get the horizontal accerleration.
ahorizontal = a * cos 22
ahorizontal = 11.07*.927 = 10.27 m/s^2

(b)
v=a*t
v=10.27m/s^2*.32s
v= 3.29 m/s (not this velocity is horizontal, assuming he didn't begin flying)
Tenelen
#4 Sep 15 2009 05:45pm
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Quote (Terminsel @ Tue, Sep 15 2009, 11:35pm)
(a)
F = m * a
m = 65kg
F = 720
a = 11.07 m/s^2
Since its at a 22 degree angle we want the cosine of 22 * 11.07 to get the horizontal accerleration.
ahorizontal = a * cos 22
ahorizontal = 11.07*.927 = 10.27 m/s^2

(b)
v=a*t
v=10.27m/s^2*.32s
v= 3.29 m/s (not this velocity is horizontal, assuming he didn't begin flying)


Alright cool that confirms it for sure, I have that as well, I just wanted to make sure I had to use cos and not sin. Thank you both for your help, I'll have another question posted in a minute if you want to make some more fg. (or just help me out, haha)


QUESTION 2:

Suppose that you are standing on a train accelerating at .20g. What minimum coefficient of static friction must exist between your feet and the floor if you are not to slide?

I have no idea how to do this type of problem. I think we are going over this tomorrow, (the work isn't due for a few more days) but I want to get it done and figure this out while I have time.

I'm assuming the goal is to get everything to equal 0 correct?

This post was edited by Tenelen on Sep 15 2009 05:50pm
Terminsel
#5 Sep 15 2009 05:53pm
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hmm be much easier to draw a diagram, but the answer is mu = 0.2

Essentially you have some mass m and a horizontal force F = ma (a = 0.2*g)
Gravity is exerting a force of mg downward on the object
the force friction exerts is mu*m*g

so you have the equation

mu*m*g = m * a
a = 0.2g
divide by m
solve for mu
mu = 0.2

oops editted forgot g somewhere.

This post was edited by Terminsel on Sep 15 2009 05:55pm
ToujoursLaVerite
#6 Sep 15 2009 05:53pm
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For this problem, think about what you know.


Your weight is equal to the normal force, since you are neither accelerating up nor down.

W = m*g = 9.8*m = Fnormal

Then you know Ffriction = Fapplied since you aren't sliding.

This means Mu * Fnormal = Ff = Fa

9.8*m*Mu = m*.20g
The masses cancel out, and you are left with Mu = (.20g)/9.8

Mu = .20

To check your answer, just know that Mu can never be greater than 1.
ToujoursLaVerite
#7 Sep 15 2009 05:54pm
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Quote (Terminsel @ Tue, Sep 15 2009, 06:53pm)
hmm be much easier to draw a diagram, but the answer is mu = 0.2/g

Essentially you have some mass m and a horizontal force F = ma (a = 0.2)
Gravity is exerting a force of mg downward on the object
the force friction exerts is mu*m*g

so you have the equation

mu*m*g = m * a
divide by m
solve for mu
mu = a/g


This is correct, knowing that a = 0.20 * g
Terminsel
#8 Sep 15 2009 05:56pm
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Nice Avatar toujours, are you a tiger?
ToujoursLaVerite
#9 Sep 15 2009 05:59pm
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Yes sir :D
Tenelen
#10 Sep 15 2009 06:02pm
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Alright....I think I understand why.

You guys are killing me, posting both the right answers at like the same time haha, I can't not pay you both.

Thanks so much for the help guys, you'll probably be seeing me around here occasionally asking for more help.
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