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Old Dec 4th 2017, 05:09 PM   #1
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Post Circular motion question URGENT

A skateboarder is trying to take a sharp turn (radius of 2.8m).He grabs a tetherball to help make the turn. As he rounds the curve, the tension in the rope is 75N and the angle the rope makes to the vertical is 28 degrees.If he makes the turn at a maximum velocity of 2.4m/s what is the coefficient of friction between the wheels and ground.The mass of the boy and board together is 45kg


I tried solving this with Fnety = Fty + Fn
but i keep getting answer wrong

Ftx=35.1104

I guess I don't know how to draw the free body diagram because I keep getting it wrong

Kudos to anyone who can solve this , i have a very important test on this tommorow

Last edited by Anon123; Dec 4th 2017 at 07:46 PM.
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Old Dec 5th 2017, 06:25 AM   #2
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centripetal force for the turn is provided by the sum of friction and the horizontal component of tension in the rope (see attached FBD)

$f + T_x = \dfrac{mv^2}{r}$

$f = \mu \cdot N = \mu(mg-T_y)$

you should be able to solve for $\mu$ from this point.
Attached Thumbnails
Circular motion question URGENT-53ebc6f4-589c-4900-b139-33dcca15d92d.png  

Last edited by skeeter; Dec 5th 2017 at 07:10 AM.
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