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Old Mar 23rd 2009, 09:43 PM   #1
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Series Of Basic Questions

Just found out that this is due in 11 hours so I'm rushing to finish, and this class is about the only one I truly do not understand, even if it is just basics.

1) Atwood Pulley m1=33kg m2=11kg
a)Determine acceleration.
b)What is the tension in the connecting string?


2) Three blocks are pulled by a force of 16.5N along a frictionless surface. What is the acceleration of the system and the tension in the string between m2 and m3?
m1=3.3kg
m2=1.1kg
m3=2.2kg


3) A boy pulls on a mass 30kg with a force of 25N.
a) Ignoring friction what is the accelaration of the box?
b) What is the normal force exerted on the box by the ground? (pulls at an angle of 30 degrees from the ground)


4) An automobile initially moving at 72 km/hr jams the brakes and skids to a stop in a distance of 24 meters on a dry pavement but stops in 81.6meters for a wet pavement(weighs 2211 lbs)
a) What is the coefficient of friction for dry weather?
b) What is the coefficient of friction for wet weather?
c) Is this static or kinetic friction?

Thanks for any help, the better you explain would be best so I can further understand the subject. I'm quite horrible in this class.
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Old Mar 23rd 2009, 09:52 PM   #2
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3) A boy pulls on a mass 30kg with a force of 25N.
a) Ignoring friction what is the accelaration of the box?
b) What is the normal force exerted on the box by the ground? (pulls at an angle of 30 degrees from the ground)

The accln of the box ignoring friction is a = F/m = 25/30 = 0.833 m/sec^2

The normal force exerted by the box on the ground is its weight - the vertical component of the force exerted = 30 x 9.6 - 25 sin 30

= 288-12.5 = 275.5 N
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Old Mar 23rd 2009, 10:05 PM   #3
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4) An automobile initially moving at 72 km/hr jams the brakes and skids to a stop in a distance of 24 meters on a dry pavement but stops in 81.6meters for a wet pavement(weighs 2211 lbs)
a) What is the coefficient of friction for dry weather?
b) What is the coefficient of friction for wet weather?
c) Is this static or kinetic friction?

The initial K.E. of 0.5 x m x (20x20) = Kd x m x 9.8 x 24 (work done against force of friction)

or Kd = 0.85 for dry Kd coeff of dynamic friction.

THe initiall K.E is used up int work against friction
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Old Mar 23rd 2009, 10:10 PM   #4
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Thanks for the help, once they are all completed I will give you the thanks thing for it. I don't know if it closes the thread once I do the thanks so I want to make sure i get the answers before I do that lol. Also does the "donation" thing go to the person getting thanks or the site in general?
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Old Mar 23rd 2009, 10:34 PM   #5
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The force on the first block is F-T1 ; the accln a = (F-T1)/m1

The force on the second block is T1-T2; the accln a = (T1-T2)/m2

The force on the third block is T2; the accln a = T2 /m3

Ti is the tension between m1 and m2, and T2 is the force of tension between m2 and m3. Solve for your answer.
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Old Mar 23rd 2009, 10:41 PM   #6
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1) Atwood Pulley m1=33kg m2=11kg
a)Determine acceleration.
b)What is the tension in the connecting string?

a = (m2-m1)g/(m1+m2) where m2 is the larger mass.

T = m1g + m1a.

Google atwoods machine youll get much better explanations than can be depicted here.
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