What is the Formula for Calculating Force of Friction on an Inclined Surface?

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SUMMARY

The discussion centers on calculating the force of friction for a box sliding down a 40.00-degree inclined surface. The box has a weight of 435 N, which translates to a mass of 44.3 kg using the formula m = Fw/g, where g is 9.81 m/s². The parallel force acting on the box is calculated as Fp = Fw(sin40) = 280 N. The net unbalanced force responsible for the box's acceleration is determined using F = m a = 11.1 N. The force of friction is then derived as Ff = 280 N - 11.1 N = 268.9 N, although there is a suggestion of a potential typo in the expected answer.

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  • Familiarity with the concept of weight and mass conversion
  • Basic principles of friction and inclined planes
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For the question: A 435 n box is sliding down a 40.00 inclined. If the acceleration of the box is 0.250m/s2. Which formula would I use to find just the force of friction? and would I change the Newtons to kilograms?
 
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The mass of the box is m = Fw/g = 435n / 9.81 m/s2 = 44.3 kg.
Is it 40 degrees?
The parallel force Fp = Fw(sin40) = 280n
The net unbalanced force responsible for acceleration is F = m a = 44.3kg (0.25m/s/s) = 11.1n.
If Ff = |Fp| then the box would not accelerate.
The Force of friction, Ff = 280n - 11.1n = 268.9n I think, but correct me if I am wrong.
Thanks.
 
Well, the answer is supposed to be 10 but I think there was probably a typo. :)
thanks so much for your help!
and yes, it was supposed to be 40.0 degrees. :)
 

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