What Minimum Force Is Needed to Prevent a Block from Sliding on a Vertical Wall?

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SUMMARY

The minimum force required to prevent a block weighing 99.0 N from sliding down a vertical wall, given a coefficient of static friction of 0.510, is calculated using the equation F = μsN. This results in a force of 50.49 N. To initiate upward movement of the block, the required force increases to 149.49 N, factoring in both the static friction and the weight of the block. These calculations are essential for understanding the dynamics of friction in vertical applications.

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  • Familiarity with Newton's laws of motion
  • Ability to perform calculations involving forces and weights
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Can someone please help me with this problem? I've tried a few different things but none of them worked...
The weight of the block in the drawing is 99.0 N. The coefficient of static friction between the block and the vertical wall is 0.510.
(a) What minimum force F is required to prevent the block from sliding down the wall? (Hint: The static frictional force exerted on the block is directed upward, parallel to the wall.)
__________ N
(b) What minimum force is required to start the block moving up the wall? (Hint: The static frictional force is now directed down the wall.)
__________ N
 

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For the first one, you'll need to use geometry to break the force into components - one horizontal, one vertical. Write an equation for the friction force using the horizontal force, add the vertical force, and set it equal to the weight of the block. Then solve for the force.

The second one is similar, but you'll need to rearrange the equation a little...
 


Sure, I'd be happy to help you with this problem! It sounds like you're on the right track by considering the coefficient of static friction and the direction of the forces involved.

For part (a), we can use the equation F = μsN, where μs is the coefficient of static friction and N is the normal force. In this case, the normal force is equal to the weight of the block, so we can plug in the given values to get F = 0.510 x 99.0 N = 50.49 N. Therefore, a minimum force of 50.49 N is required to prevent the block from sliding down the wall.

For part (b), we can use the same equation but with the force of gravity acting in the opposite direction. So, F = μsN + mg, where mg is the force of gravity pulling the block down. Again, plugging in the given values, we get F = 0.510 x 99.0 N + 99.0 N = 99.0 N + 50.49 N = 149.49 N. Therefore, a minimum force of 149.49 N is required to start the block moving up the wall.

I hope this helps! Remember to always double check your calculations and units to make sure your answer is correct. Best of luck!
 

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