Calculating Fmax and Fmin for an inclined block with static friction

In summary, the block is being pushed up by a force of 8.9 Newtons, but if the force were applied in the parallel dimension of the gravitational force, the maximum force would be 37.5 Newtons.
  • #1
TLeo198
8
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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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  • #2
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?
 
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  • #3
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.
 
  • #4
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
 
  • #5
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.
 

FAQ: Calculating Fmax and Fmin for an inclined block with static friction

What is static friction?

Static friction is a type of force that prevents two surfaces from sliding against each other when they are not in motion.

How is static friction different from kinetic friction?

Static friction occurs when two surfaces are not moving against each other, while kinetic friction occurs when two surfaces are already in motion.

Why is static friction important?

Static friction plays a crucial role in everyday life, as it allows objects to remain stationary on a surface, such as a book on a table. It also helps with traction and prevents slipping.

How is static friction calculated?

The force of static friction can be calculated by multiplying the coefficient of static friction between two surfaces by the normal force, which is the force exerted by one surface on another.

Can the force of static friction be greater than the force applied?

Yes, the force of static friction can be greater than the force applied, up to a certain limit known as the maximum static friction force. Once this limit is reached, the object will begin to move and kinetic friction takes over.

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