# How much force would the man apply, along the handle, if the coefficient of friction were .128?

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Mass of lawnmower=75 kg

Angle of handle to horizontal: 37 degrees

Force along handle without friction: 93.91 newtons

Horizontal force component: Fcos37

Vertical Force Component: Fsin37

Acceleration: 1 m/s^2

Apr 5th, 2015

F cos 37 - uk (mg+mgsin37)  = ma

F = 75+0.128(75*9.8+75*9.8*0.6)

=225.53N

Apr 5th, 2015

Thank you!

Apr 5th, 2015

Could you tell me why it's mgsin37, and not 93.91sin37?

Apr 5th, 2015

maybe I understand the question wrong, I thought we are supposed to find the force. with friction, I remember its has friction of 0.128, but now can;t see it anywhere in the problem

Apr 5th, 2015

Isn't the formula something like this: Horizontal force=Force on handle minus the force of friction, where the force of friction is the coefficient times the sum of the vertical component of the force and mg?

Apr 5th, 2015

yes the sum is mg sin37 + mg

Apr 5th, 2015

So replacing F with mg in Fsin37 is appropriate in this context? Because I thought F was the force on the handle.

Apr 5th, 2015

If F is given, then what we are suppose to find in the problem

Apr 5th, 2015

So since the sum of those two is the normal force, we use mg?

Apr 5th, 2015

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Apr 5th, 2015
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Apr 5th, 2015
Nov 17th, 2017
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