friction: finding coefficient from mass and force

[Guest]

Does anyone fancy having a go at this?

A box containing engine components has a combined mass of 450 kilograms. The box is in a warehouse having a smooth concrete floor. The force required to move this unit up a slope of 30 is 3.6 kN. Determine the coefficient of friction between the floor and the box.

 

[stapel]

Is the slope 30 "degrees", or some other unit? What formulas have you been given? How far have you gotten in applying them?

Please reply with specifics. Thank you.

Eliz.

 

[skeeter]

moving the box up an incline at a constant speed would require an applied force up the incline equal to the box's component of weight down the incline plus the force of kinetic friction acting on the box, which is also acting down the incline.

\(\displaystyle \L F = mgsin(\theta) + \mu_k mgcos(\theta)\)

 

[Guest]

Yes, the slope is 30 "degrees". I have been given no formulas.

Thanks