Solving Friction Force and Coefficient on a 17 kg Box on a 35 Degree Incline

AI Thread Summary
To solve for the friction force on a 17 kg box on a 35-degree incline, the equation F - F_friction = ma is used, where F is the gravitational force component down the incline. The calculated friction force is approximately 91N, but this value did not yield a correct answer in the webassign program. A suggested friction force of 92.9N was provided as a potential solution. The coefficient of friction remains to be calculated based on the correct friction force.
shrtweez13
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i've been trying to figure this question out for about half an hour now and I'm still where I started. if some one could help me i'd be much obliged.

a 17 kg box is released on a 35 degree incline and accelerates down the incline at .270 m/s^2. Find the friction force impeding its motion. How large is the coefficient of friction?
 
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F-Ffriction=ma
(17kg*9.8m/s^2*sin35)-Ff=17kg*.27m/s^2
Ff=91N
 
i kind of understand what you did but its not the answer. I'm using the webassign program that let's you check your homework and i tried that answer and it didn't turn out right.
 
what answer did you plug in? Try 91N, 92N, or 93N

I'm leaving the coefficient of friction for you to solve
 
92.9N.
I got this
 
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