Calculating Fmax and Fmin for an inclined block with static friction

AI Thread Summary
The discussion revolves around calculating the minimum (Fmin) and maximum (Fmax) forces for a block on an incline with static friction. A user initially calculated Fmax as 8.9 N using the static friction formula but received incorrect results. Participants pointed out the need to consider the gravitational force component acting parallel to the incline, which affects the calculations. The correct approach involves adding this gravitational component to the initial force calculation. The user acknowledged the misunderstanding and expressed gratitude for the clarification.
TLeo198
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I'm having trouble with this one problem, I've tried everything with my given notes, but the right answer isn't coming up:

A block with a mass of 3.7 kg is placed at rest on a surface inclined at an angle of 52 degrees above the horizontal. The coefficient of static friction between the block and the surface is 0.40, and a force of magnitude F pushes upward on the block, parallel to the inclined surface. Calculate Fmin and Fmax.

For Fmax, I got 8.9, (had to report to 2 sigfigs), by using the formula fs,max = mu(static coefficient) x N. N = mgcos(theta). My answer was wrong. Please, if anyone can help me find how to do this, it would be greatly appreciated!
 
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I think your problem is in the fact that the applied force is in the horizontal direction. Consider the fact that only the component of F along the hill can act against the static friction force. So, when considering the relationships between friction and the maximum and minimum values for F, you are only going to want to consider the component of F along the hill. Do you see how this will change the relationships you are using?
 
Last edited:
No, applied force is parallel to the slope. He forgot to take into account the component of gravitational force, that is parallel to the ground.
 
Yo forgot to add the additional force acting in the parallel dimension of the gravitational force which will be (sin(52))(36.26) and then you would add that to your 8.9 and get 37.5 Newtons
 
Lojzek said:
No, applied force is parallel to the slope. He forgot to take into account the component of gravitational force, that is parallel to the ground.

glennpagano44 said:
Yo forgot to add the additional force acting in the parallel dimension of the gravitational force which will be (sin(52))(36.26) and then you would add that to your 8.9 and get 37.5 Newtons

Wow yes, you are correct. I don't know what I was thinking here... I must have read the problem incorrectly. I'm sorry for any confusion I caused.
 

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