Asked by solexgirl12

This question came up on my review package tonight, and it has me stumped.

"A box of mass 14.5kg slides at a constant speed of 4.70m/s down a ramp at 35.3 degrees below the horizontal. What is the force of friction on the box?"

I can calcuate force of friction on accelerating objects, but one moving at a constant speed has me puzzled. Also, my FBD Fgx and Fgy components look a little suspicious. I got:
Fgy=142.1 N
Fgx=79.6 N

Can anyone help clear this up for me?
Thanks so much.
Answered by drwls
If there is no acceleration, the component of weight down the ramp, M g sin 35.3, equals the friction force.

The speed does not matter in the calculation.

I do't see how you got your Fgy and Fgx umbers.
Answered by solexgirl12
Yeah, me either. *sigh* Thanks very much.
Answered by Anonymous
yeah
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