Finding Friction force impeding boxs motion

AI Thread Summary
To find the friction force impeding the motion of a 21.0 kg box on a 36.0° incline accelerating at 0.271 m/s², the gravitational force component acting down the incline is calculated using F = mg*cos(36). This results in a force of approximately 166.495 N acting downwards. The net force equation indicates that the acceleration is the difference between the downward forces and the upward friction force. The discussion highlights confusion regarding the direction of forces, clarifying that friction opposes the motion of the box. The calculations and understanding of forces are critical for accurately determining the friction force.
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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