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Given the coefficient of friction =1.0 and the mass of the object= 1.081 kg, at what...
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You need to evaluate the forces acting on the object on the plane, such that:
`G = m*g`
`F_G = G*sin theta` (`theta` is the angle of the incline, `F_G` is the force that is parallel to the inclined plane)
`F_N = G*cos theta` (`F_N` is the component force of G force, perpendicular to the inclined plane)
The equation that relates the friction force to the normal force `F_N` is the following:
`F_f = mu*F_N => F_f = mu*G*cos theta`
Balancing the forces yields:
`G*sin theta = mu*G*cos theta `
Reducing duplicate factors both sides yields:
`sin theta = mu*cos theta => sin theta/cos theta = mu`
The problem provides the coefficient of friction `mu` , such that:
`sin theta/cos theta = 1 => tan theta = 1 => theta = pi/4`
Hence, evaluating the angle of the incline at object will slide down on plane, yields `theta = pi/4` .
Posted by sciencesolve on May 17, 2013 at 5:10 PM (Answer #3)
I got it from somewhere. Fnet has to be greater than 0 for it to move, right? This is called the critical angle.
`F = ma> 0`
` ` `upsilon = 1.0 `
F parallel=F friction
(m)(g)sin `theta = (upsilon)(m)(g)cos theta`
You'll end up with:
sin theta/cos theta = 1
And this is equal to:
theta=inverse tan 1
theta = 45 degrees
Posted by edelquinnparacuelles on April 7, 2013 at 1:07 PM (Answer #2)
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