Normal contact force for a sliding slope

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For the following question, how do we know that the acceleration of ##M## is constant over time? And is the normal contact force between the two masses smaller as compared to that where ##M## is fixed?

Screen Shot 2016-06-20 at 4.34.29 pm.png


The acceleration depends on the net force on ##M##, which depends on the normal contact force ##N## between ##m## and ##M##. The force ##N## depends on how tightly the two surfaces are pressed together. So it seems plausible that ##N## is smaller when ##M## is free to slide compared to that when ##M## is fixed.

The initial velocities of ##m## and ##M## are zero. Suppose we set up the initial conditions by holding ##m## and ##M## fixed and then release them. Would the initial (when both masses are just released from grip) value of ##N## in this case be the same as the value of ##N## where ##M## is always fixed?

I guess they would be the same. But since the acceleration is constant over time, the value of ##N##, at all other time after ##M## starts to slide, must be the same as the initial value of ##N##. Then, we must conclude that the value of ##N## where ##M## is fixed is equal to the value of ##N## where ##M## is sliding away from ##m##. But when ##M## is sliding, it seems that the two masses are less tightly pressed together, and so ##N## should be smaller.
 
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  • #2
wrobel
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you should write the equations of motion to this system.
 
  • #3
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you should write the equations of motion to this system.
I get ##N## is constant provided ##a_x## or ##A_x## or ##N'## is constant, where ##a## and ##A## are the accelerations of ##m## and ##M##, and ##N'## is the normal contact force on ##M## by the floor.

How do we know ##a_x## or ##A_x## or ##N'## is constant?

Suppose ##N## is velocity dependent. I believe the equations of motion, eg., ##N\cos\theta+Mg=N'##, still holds, just that now ##a_x##, ##A_x## and ##N'## will be time dependent. So it doesn't seem like we can deduce ##N## is time independent from the equations of motion.
 
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  • #4
wrobel
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047f1252641b.png


$$m\boldsymbol a_m=\boldsymbol N+m\boldsymbol g,\quad M\boldsymbol a_M=-\boldsymbol N+\boldsymbol R+M\boldsymbol g;\qquad (*)$$
$$\boldsymbol a_m=\boldsymbol a_M+w\boldsymbol e_x;\quad \boldsymbol N=N\boldsymbol e_y;$$
$$\boldsymbol n=\cos\theta\boldsymbol e_y-\sin\theta\boldsymbol e_x;\quad \boldsymbol e=\sin\theta \boldsymbol e_y+\cos\theta\boldsymbol e_x;$$
$$\boldsymbol a_M=a_M\boldsymbol e,\quad \boldsymbol R=R\boldsymbol n,\quad \boldsymbol g=-g\boldsymbol n$$
so we have four unknowns ##w,a_M,R,N## and four scalar equations (*)
 
  • #5
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047f1252641b.png


$$m\boldsymbol a_m=\boldsymbol N+m\boldsymbol g,\quad M\boldsymbol a_M=-\boldsymbol N+\boldsymbol R+M\boldsymbol g;\qquad (*)$$
$$\boldsymbol a_m=\boldsymbol a_M+w\boldsymbol e_x;\quad \boldsymbol N=N\boldsymbol e_y;$$
$$\boldsymbol n=\cos\theta\boldsymbol e_y-\sin\theta\boldsymbol e_x;\quad \boldsymbol e=\sin\theta \boldsymbol e_y+\cos\theta\boldsymbol e_x;$$
$$\boldsymbol a_M=a_M\boldsymbol e,\quad \boldsymbol R=R\boldsymbol n,\quad \boldsymbol g=-g\boldsymbol n$$
so we have four unknowns ##w,a_M,R,N## and four scalar equations (*)
There should be 5 unknowns because ##m## could accelerate vertically as it slides down the slope.

Anyway, my question is how do we know the terms in the equations of motion are time independent?
 
  • #6
wrobel
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mm could accelerate vertically
sure, that has already been taken into account, look at the formulas carefully
 
  • #7
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sure, that has already been taken into account, look at the formulas carefully
Shouldn't it be ##\mathbf {a_m}=\mathbf {a_M}+w\mathbf {e_x}+u\mathbf {e_y}##?

##M## is free to slide too. It accelerates to the right.
 
  • #8
wrobel
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the term ##w\boldsymbol e_x## is the acceleration of ##m## relative to the wedge; ##\boldsymbol a_M## is acceleration of the wedge. We use the summation of accelerations theorem, the term ##u\boldsymbol e_y## is unnecessary
 
  • #9
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the term ##w\boldsymbol e_x## is the acceleration of ##m## relative to the wedge; ##\boldsymbol a_M## is acceleration of the wedge. We use the summation of accelerations theorem, the term ##u\boldsymbol e_y## is unnecessary
What's the theorem about? The relative acceleration won't just be in the x direction, so I don't see why you only introduce a constant ##w##.

Anyway, this doesn't answer the question of time independence.
 
  • #10
wrobel
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you do not understand banal things but what a self-confidence :)
 
  • #11
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you do not understand banal things but what a self-confidence :)
You made a mistake isn't it? There should be 5 unknowns.
 
  • #12
wrobel
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this continues posting #4. Solving linear system (*) we get
$$w=cg(m+M)\sin\theta,\quad a_M=-\frac{c}{2}mg\sin 2\theta,\quad R=cMg(M+m),\quad N=c mMg\cos\theta, $$
here $$c=\frac{1}{M+m\sin^2\theta}.$$ Calculations are in the attachment


Simpler way to solve this problem is to use the Lagrange equations. There is another problem of the same type but little bit more complicated : to replace the block with a disk that rolls on the wedge without slipping.
 

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