# I Normal contact force for a sliding slope

1. Jun 20, 2016

### Happiness

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?

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.

Last edited: Jun 20, 2016
2. Jun 20, 2016

### wrobel

you should write the equations of motion to this system.

3. Jun 20, 2016

### Happiness

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.

Last edited: Jun 20, 2016
4. Jun 20, 2016

### wrobel

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

5. Jun 20, 2016

### Happiness

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. Jun 20, 2016

### wrobel

sure, that has already been taken into account, look at the formulas carefully

7. Jun 20, 2016

### Happiness

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. Jun 20, 2016

### wrobel

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. Jun 20, 2016

### Happiness

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. Jun 20, 2016

### wrobel

you do not understand banal things but what a self-confidence :)

11. Jun 20, 2016

### Happiness

You made a mistake isn't it? There should be 5 unknowns.

12. Jun 21, 2016

### wrobel

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.

#### Attached Files:

• ###### wedge and block.pdf
File size:
223.6 KB
Views:
83
Last edited: Jun 21, 2016