Calculating Friction Force on 18.8kg Box on 38° Incline

[p0ink]

An 18.8kg box is released on a 38.0o incline and accelerates down the incline at 0.281m/s2. What is the magnitude of the friction force impeding its motion.

F = m*g*cos(38)
sum forces parallel to the plane
m*g*sin(38) - (mu)*(F) = m*a, or
m*g*sin(38) - (mu)*m*g*cos(38) = m*a
masses cancel out
[9.81*sin(38)-(.281)]/[9.81*cos(38)] = .75 = mu

but it says I'm wrong. i can't be.

i only have one try left, and that's it.

 

[Doc Al]

All you are asked to find is the friction force. No need to compute the normal force (what you call F, for some reason) or find mu. Rewrite your equation for forces parallel to the plane using "F" to represent the friction force.

 

[kyle8921]

I understand your frustration with only having one try left to solve this problem. However, it is important to approach this situation with a calm and rational mindset. It is possible that there may be a mistake in your calculations or a misunderstanding of the problem. I would suggest double-checking your calculations and the given values to ensure accuracy. Additionally, it may be helpful to consult with a colleague or reference materials to verify your approach and calculations. It is also important to remember that mistakes and challenges are a natural part of the scientific process and can lead to valuable learning experiences. Keep an open mind and continue to approach the problem with determination and critical thinking. Good luck!