Finding Friction force impeding boxs motion

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SUMMARY

The discussion focuses on calculating the friction force acting on a 21.0 kg box on a 36.0° incline, which accelerates downwards at 0.271 m/s². The gravitational force component acting parallel to the incline is calculated using the formula F = mg*cos(θ), resulting in a force of approximately 166.495 N. The user expresses confusion about the forces acting on the box, specifically the relationship between the weight component and friction. The correct approach involves setting the net force equal to the product of mass and acceleration.

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cleo0724
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A 21.0 kg box is released on a 36.0° incline and accelerates down the incline at 0.271 m/s2. Find the friction force impeding its motion.

F=mg*cos(36)

F=21(9.8)*cos(36)
F=205.8*.809
F=166.495

I'm not sure what I'm doing wrong. Please help
 
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the component of the weight acts downwards and the friction acts upwards (directions are parallel to the plane)


so if the box moves down then ma=forces down-forces up.


a bit easier now?
 

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