A block of mass m, acted on by a force of magnitude F directed horizontally to the right as shown above, slides up an inclined plane that makes an angle [tex]\theta[/tex] with the horizontal. The coefficient of sliding friction between the block and the plane is [tex]\mu[/tex].(adsbygoogle = window.adsbygoogle || []).push({});

The picture is that of an incline, [tex]\theta[/tex] on the left of the incline, mass m on the incline, and a horizontal force is applied to m directed right.

Develop an expression for the magnitude of the force F that will allow the block to slide up the plane with constant velocity. What relation must [tex]\theta[/tex] and [tex]\mu[/tex] satisfy in order for this solution to be physically meaningful.

This is what I did. (wrong according to the answer)

F = ma

mgsin[tex]\theta[/tex] + [tex]\mu[/tex]mgcos[tex]\theta[/tex] - Fcos[tex]\theta[/tex] = 0

Fcos[tex]\theta[/tex] = mg(sin[tex]\theta[/tex] + [tex]\mu[/tex]cos[tex]\theta[/tex])

F = mg(sin[tex]\theta[/tex] + cos[tex]\theta[/tex]) / cos[tex]\theta[/tex]

This is how I reasoned: If it's moving up at a constant velocity, Fcos[tex]\theta[/tex] should equal mgsin[tex]\theta[/tex] + friction ([tex]\mu[/tex]gcos[tex]\theta[/tex])

This is what the answer is supposed to be:

F = mg([tex]\mu[/tex]cos[tex]\theta[/tex] + sin[tex]\theta[/tex]) / (cos[tex]\theta[/tex] - [tex]\mu[/tex]sin[tex]\theta[/tex])

F > 0 => cos[tex]\theta[/tex] > sin[tex]\theta[/tex]

tan[tex]\theta[/tex] = 1/[tex]\mu[/tex]

What mistake did I make in my reasoning and where did the answer come from?

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# Homework Help: A problem that has force, friction on an incline

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