# Describe how the equilibrium is broken (mechanics)

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#1
Question (c): http://prntscr.com/ugutvc

This question doesn't make much sense. Box A is in equilibirum so why are they telling me the maximum friction force possible between it and the ground? If there is a net force forwards/backwards on box B then box A should stay still, right?

Based on the way it is worded I'm assuming they want me to treat box A and B as 1 particle like I did in parts a and b so I can try to solve the 2 equations simultaneously to find k. But that is obviously wrong as I end up with something like this: http://prntscr.com/ugvdww
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1 week ago
#2
There are two constraints.
One is for the single box net force being <= 120 (max possible friction)
The other for both boxes net force being <= 180
The question is asking if you increase k, which constraint breaks first. You could find the value of k where the net force equals the max frictional force in each case, then take the smallest k to demonstrate which breaks first.
Last edited by mqb2766; 1 week ago
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#3
(Original post by mqb2766)
There are two constraints.
One is for the single box net force being <= 120 (max possible friction)
The other for both boxes net force being <= 180
The question is asking if you increase k, which constraint breaks first. You could find the value of k where the net force equals the max frictional force in each case, then take the smallest k to demonstrate which breaks first.
But it just says box A and the ground not box A and B with the ground which means you're not suppose to combine them unless their shape and sizes are the same. If box B has a greater net force then wouldn't it just slide off box A which remains still? If you combine them into 1 box then there is only 1 constraint which breaks so you can't compare them....
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1 week ago
#4
(Original post by TSR360)
But it just says box A and the ground not box A and B with the ground which means you're not suppose to combine them unless their shape and sizes are the same. If box B has a greater net force then wouldn't it just slide off box A which remains still? If you combine them into 1 box then there is only 1 constraint which breaks so you can't compare them....
If the box A -box B friction is exceeded, box B slides away from A and you treat as seperate.
Otherwise if the friction is not exceeded, you treat them as a single object and check whether A (and B) exceeds the A - ground frictiion.
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