The minimum applied force before an object slips off wall

AI Thread Summary
A horizontal force is applied to hold a block against a vertical wall, with the block's mass being 1.7 kg and coefficients of static and kinetic friction at 0.87 and 0.81, respectively. The block will slip when the weight exceeds the frictional force, leading to the calculation of the normal force, which is approximately 19 N. The total contact force is a combination of the frictional and normal forces, requiring vector addition to determine its magnitude and direction. The angle of the contact force with the horizontal is essential for understanding the forces at play, with discussions suggesting it could be 0 or 90 degrees depending on the interpretation of the forces involved. Ultimately, the problem requires a clear understanding of both the magnitude and direction of the total contact force just before the block slips.
blueray101
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Homework Statement


A horizontal force F is applied to hold a block, mass m = 1.7 kg against a vertical wall. The coefficients of kinetic and static friction between the block and wall are 0.81 and 0.87 respectively. F is gradually reduced until the block falls. Just before it falls, what is the magnitude of the total contact force Fc exerted by the block on the wall, and what is the angle θ it makes with horizontal? (+ve for above, -ve for below the horizontal).

Homework Equations


μsN=Ff
F=mg

The Attempt at a Solution



So, if weight surpasses friction, it slips.

μsN<W

N<W/μs

N=1.7*9.8/0.87=19N approximately

I have no clue what it is referring to when it asks me to find the angle. The contact force seems to be horizontal so angle is 0?
 
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blueray101 said:

Homework Statement


A horizontal force F is applied to hold a block, mass m = 1.7 kg against a vertical wall. The coefficients of kinetic and static friction between the block and wall are 0.81 and 0.87 respectively. F is gradually reduced until the block falls. Just before it falls, what is the magnitude of the total contact force Fc exerted by the block on the wall, and what is the angle θ it makes with horizontal? (+ve for above, -ve for below the horizontal).

Homework Equations


μsN=Ff
F=mg

The Attempt at a Solution



So, if weight surpasses friction, it slips.

μsN<W

N<W/μs

N=1.7*9.8/0.87=19N approximately

I have no clue what it is referring to when it asks me to find the angle. The contact force seems to be horizontal so angle is 0?
The total contact force consists of the frictional force and the normal force.
 
haruspex said:
The total contact force consists of the frictional force and the normal force.
Is this directed at the 2nd or 1st part of the question? I assume the first part is correct then?

If the contact force is vertical, then the angle it makes with the horizontal is 90 degrees?
 
blueray101 said:
Is this directed at the 2nd or 1st part of the question? I assume the first part is correct then?

If the contact force is vertical, then the angle it makes with the horizontal is 90 degrees?
You need to find the angle of the net force. 2 Forces are considered contact forces: Friction and normal force here that is what he meant
 
blueray101 said:
Is this directed at the 2nd or 1st part of the question? I assume the first part is correct then?
Both parts. You have to think of the normal force and frictional force as adding up, vectorially, to a single contact force. The question is aksing for its magnitude snd direction.
 

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