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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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edelquinnpara... | Student, Grade 11 | eNotes Newbie

Posted April 7, 2013 at 11:12 AM via web

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Given the coefficient of friction =1.0 and the mass of the object= 1.081 kg, at what incline angle would the object have started moving?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted May 17, 2013 at 5:10 PM (Answer #3)

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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` .

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edelquinnparacuelles | Student , Grade 11 | eNotes Newbie

Posted April 7, 2013 at 1:07 PM (Answer #2)

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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 `

`m=1.081 kg`

F parallel=F friction

(m)(g)sin `theta = (upsilon)(m)(g)cos theta`

(Fg cancels)

You'll end up with: 

sin theta/cos theta = 1

And this is equal to:

tan theta=1

theta=inverse tan 1

theta = 45 degrees

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