Maximum friction force before slipping occurs

AI Thread Summary
The discussion centers on understanding the maximum friction force before slipping occurs, defined by the equation f_s ≤ μ_sN, where μ_s is the coefficient of static friction and N is the normal force. Participants clarify that slipping happens when the applied force exceeds this maximum frictional force. A practical example is provided involving a block on an inclined slope, illustrating that the block will slide when the gravitational force component parallel to the slope surpasses the maximum frictional force. Additionally, a scenario is introduced involving a crate on a truck climbing a hill, prompting further exploration of calculating maximum acceleration without slipping. The conversation emphasizes the importance of grasping the relationship between friction, normal force, and external forces to prevent slipping.
electricblue
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Homework Statement



Hi guys, i am sort of confused with the concept of slipping. I got a question.. what does it mean by maximum friction force before slipping occurs?

Homework Equations



I believe in the use of f_{}s \leq \mu_{}sn

The Attempt at a Solution



please correct me if i am wrong. maximum friction force before slipping occurs = \mu_{}sn
 
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Hi electricblue! :smile:

(have a mu: µ and a ≤ and try using the X2 tag just above the Reply box :wink:)
electricblue said:
Hi guys, i am sort of confused with the concept of slipping. I got a question.. what does it mean by maximum friction force before slipping occurs?
…
please correct me if i am wrong. maximum friction force before slipping occurs = \mu_{}sn

Yes, if the force needed was any more (than µsN), then the force available wouldn't be enough, and so there would be slipping. :smile:
 
Hi electricblue,

To make you understand this concept, consider a block placed on an angled slope.

The block can either;

1. Slide (the frictional force exerted on the slope by the contact surface of the object < the force component due to gravity acting on the block parallel to the slope).

2. Remain stationary (the frictional force exerted on the slope by the contact surface of the object >= the force component due to gravity acting on the block parallel to the slope).

Let's say that the block originally remains stationary.

Now let's increase the angle of the slope. This has the same effect as to increase the force component due to gravity acting on the block parallel to the slope (You may find this easier to conceptualise if you draw a diagram).

There will come a point when the block begins to slide, that is, when the maximum frictional force between the contact surface and the block is overcome. This maximum frictional force is given by the equation you've stated, in words:

Maximum frictional force = coefficient of friction x normal force
 
Ok thanks! I got it now
 
I have been studying this concept of slipping now it gets me pondered..

what if i have a crate on a truck and the truck is suppose to climb up a hill of 20 degrees. mass of crate is 100kg and coefficent of static friction and kinetic friction is 0.4 and 0.5 respectively. the question is: What is the maximum acceleration the truck can have without the crate slipping?
 
Hi again electricblue,

You're going to have to show some attempt before we can help you.

Hint: Use N2L and also work out the maximum frictional force before slipping.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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