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Author Topic: motion Q  (Read 678 times)  Share 

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TrueTears

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motion Q
« on: November 24, 2008, 04:01:17 pm »
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A cart of mass 250g sliding down a plane inclined at 25 degeree to the horizontal with uniform speed. A mass of 50 g hangs over a pully and provides a tension force on the cart. Find the normal reaction force(Fn), the weight of the cart(Fw ), the component of the cart's weight down the slope in the direction of motion(Fx) and the friction force on the cart(Ff). i found the 1st 3 ( which are all correct according to the answers), Fn = 2.22N Fw = 2.45N Fx = 1.035N but Ff = ?
how to work out friction force?

Well, i tried this way the total Force = Tension force (provided by the weight) + F x ( the component of the cart's weight down the slope in the direction of motion) - Friction force.

So total force = ma = m x gsin25= ((250+50)/1000) x 9.8 x sin25 = 1.24
Tension force = (50/1000) x 9.8
F x = 1.035
so 1.24 = (50/1000) x 9.8  + 1.035 - Ff
Ff = 0.28N however this is wrong. why? and what's the correct working?
« Last Edit: November 24, 2008, 04:10:43 pm by TrueTears »
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Re: motion Q
« Reply #1 on: November 24, 2008, 05:12:45 pm »
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not sure if it's right lol
« Last Edit: November 24, 2008, 05:14:16 pm by DivideBy0 »