# Unfixed Mass and Wedge - am I correct?

1. May 3, 2008

### Ataman

1. The problem statement, all variables and given/known data

A mass m1 sits on an incline with mass m2 and angle $$\theta$$. Find the accelerations of m1 and m2 relative to the ground after they are released.

This is a long and tricky problem (since we do not know the normal forces, and we have to solve for a combinations of accelerations). Therefore I would like people to check if this is correct.

2. Relevant equations

$$\vec{F_{net}} = m \vec{a}$$
$$\vec{a}_{_{2G}}+\vec{a}_{_{12}}=\vec{a}_{_{1G}}$$

In other words, acceleration of m1 relative to the ground equals acceleration of m2 relative to the ground plus acceleration of m1 relative to m2.

3. The attempt at a solution

Let +i point right (the opposite direction the acceleration of the wedge), and +j point down for all systems.

Let $$\vec{F}_{_{N1}}$$ be the normal force between the block and the wedge (unknown), and let $$\vec{F}_{_{N2}}$$ be the normal force between the wedge and the ground (also unknown).

We do not know the magnitude of the acceleration of the block relative to the wedge, but we know it is along the incline, so

$$a_{_{12X}}\hat{i}=\left|\vec{a}_{_{12}}\right|cos\theta$$
$$a_{_{12Y}}\hat{j}=\left|\vec{a}_{_{12}}\right|sin\theta$$

$$\vec{F_{net}} = m_{_{1}} \vec{a}_{_{1G}}$$ system: m1
$$\vec{F_{_{G}}}+\vec{F_{_{N1}}} = m_{_{1}} \vec{a}_{_{1G}}$$

$$\hat{i}:F_{_{N1X}}=m_{_{1}}a_{_{1GX}}$$
$$\hat{i}:F_{_{N1}}sin\theta=m_{_{1}}a_{_{1GX}}$$
$$\hat{i}:F_{_{N1}}=\frac{m_{_{1}}a_{_{1GX}}}{sin\theta}$$

$$\hat{j}:F_{_{G}}-F_{_{N1Y}}=m_{_{1}}a_{_{1GY}}$$
$$\hat{j}:F_{_{G}}-F_{_{N1Y}}=m_{_{1}}a_{_{1GY}}$$
$$\hat{j}:m_{_{1}}g-F_{_{N1}}cos\theta=m_{_{1}}a_{_{1GY}}$$
$$\hat{j}:m_{_{1}}g-\left(\frac{m_{_{1}}a_{_{1GX}}}{sin\theta}\right)cos\theta=m_{_{1}}a_{_{1GY}}$$
$$\hat{j}:g-a_{_{1GX}}cot\theta=a_{_{1GY}}$$

$$\vec{F_{net}} = m_{_{2}} \vec{a}_{_{2G}}$$ system: m2
$$\vec{F_{_{G}}}+\vec{F_{_{N1}}}+\vec{F_{_{N2}}} = m_{_{2}} \vec{a}_{_{2G}}$$

$$\hat{i}:0-F_{_{N1X}}+0=-m_{_{2}}a_{_{2GX}}$$
$$\hat{i}:F_{_{N1}}sin\theta=m_{_{2}}a_{_{2G}}$$
$$\hat{i}:\left(\frac{m_{_{1}}a_{_{1GX}}}{sin\theta}\right)sin\theta=m_{_{2}}a_{_{2G}}$$
$$\hat{i}:m_{_{1}}a_{_{1GX}}=m_{_{2}}a_{_{2G}}$$
$$\hat{i}:m_{_{1}}\left(a_{_{2GX}}+a_{_{12X}}\right)=m_{_{2}}a_{_{2G}}$$
$$\hat{i}:m_{_{1}}\left(a_{_{2G}}+a_{_{12}}cos\theta\right)=m_{_{2}}a_{_{2G}}$$
$$\hat{i}:m_{_{1}}a_{_{12}}cos\theta=\left(m_{_{2}}-m_{_{1}}\right)a_{_{2G}}$$
$$\hat{i}:\frac{m_{_{1}}a_{_{12}}cos\theta}{\left(m_{_{2}}-m_{_{1}}\right)}=a_{_{2G}}$$

$$\hat{j}:F_{_{G}}+F_{_{N1Y}}-F_{_{N2Y}}=m_{_{2}}a_{_{2GY}}$$
$$\hat{j}:m_{_{2}}g+F_{_{N1}}cos\theta-F_{_{N2}}=0$$
$$\hat{j}:m_{_{2}}g+F_{_{N1}}cos\theta=F_{_{N2}}$$

Now combining the unknowns to solve for the accelerations:

We know that
$$\hat{i}:a_{_{2GX}}+a_{_{12X}}=a_{_{1GX}}$$
$$\hat{i}:a_{_{2G}}+a_{_{12X}}=a_{_{1GX}}$$
$$\hat{i}:a_{_{2G}}+a_{_{12}}cos\theta=a_{_{1GX}}$$
and
$$\hat{j}:a_{_{2GY}}+a_{_{12Y}}=a_{_{1GY}}$$
$$\hat{j}:a_{_{12Y}}=a_{_{1GY}}$$
$$\hat{j}:a_{_{12}}sin\theta=a_{_{1GY}}$$

From the j component of the system m1, we know that
$$\hat{j}:g-a_{_{1GX}}cot\theta=a_{_{1GY}}$$
and we also know from above that:
$$\hat{j}:a_{_{1GY}}=a_{_{12Y}}$$
and
$$\hat{i}:a_{_{2G}}+a_{_{12}}cos\theta=a_{_{1GX}}$$

Therefore,
$$\hat{j}:g-a_{_{1GX}}cot\theta=a_{_{12Y}}$$
$$\hat{j}:g-\left(a_{_{2G}}+a_{_{12}}cos\theta\right)cot\theta=a_{_{12}}sin\theta$$

We derived for system m2 what $$a_{_{2G}}$$ was, so we can solve for $$a_{_{12}}$$
$$g-\left(\frac{m_{_{1}}a_{_{12}}cos\theta}{\left(m_{_{2}}-m_{_{1}}\right)}+a_{_{12}}cos\theta\right)cot\theta=a_{_{12}}sin\theta$$
$$g-a_{_{12}}cos\theta\left(\frac{m_{_{1}}}{m_{_{2}}-m_{_{1}}}+1\right)cot\theta=a_{_{12}}sin\theta$$
$$g=a_{_{12}}cos\theta\left(\frac{m_{_{1}}}{m_{_{2}}-m_{_{1}}}+1\right)cot\theta+a_{_{12}}sin\theta$$
$$g=a_{_{12}}\left(cos\theta\left(\frac{m_{_{1}}}{m_{_{2}}-m_{_{1}}}+1\right)cot\theta+sin\theta\right)$$

$$\frac{g}{cos\theta\left(\frac{m_{_{1}}}{m_{_{2}}-m_{_{1}}}+1\right)cot\theta+sin\theta}=a_{_{12}}$$

We finally have the magnitude of $$a_{_{12}}$$, but we know the direction, since the acceleration of the mass relative to the incline is down the incline itself.

$$\vec{a}_{_{12}}=a_{_{12X}}\hat{i}+a_{_{12Y}}\hat{j}$$
$$\vec{a}_{_{12}}=a_{_{12}}cos\theta\hat{i}+a_{_{12}}sin\theta\hat{j}$$
$$\vec{a}_{_{12}}=a_{_{12}}(cos\theta\hat{i}+sin\theta\hat{j})$$
$$\vec{a}_{_{12}}=\left(\frac{g}{cos\theta\left(\frac{m_{_{1}}}{m_{_{2}}-m_{_{1}}}+1\right)cot\theta+sin\theta}\right)(cos\theta\hat{i}+sin\theta\hat{j})$$

Now we find the acceleration of the block relative to the earth.

We were able to find an expression for the acceleration of the block relative to the ground in terms of the given variables and, most importantly, the magnitude of the acceleration of the block, which we just solved.

From the i-component of system m2:
$$\frac{m_{_{1}}a_{_{12}}cos\theta}{\left(m_{_{2}}-m_{_{1}}\right)}=a_{_{2G}}$$
we go to here...
$$\left(\frac{g}{cos\theta\left(\frac{m_{_{1}}}{m_{_{2}}-m_{_{1}}}+1\right)cot\theta+sin\theta}\right)\left(\frac{m_{_{1}}cos\theta}{\left(m_{_{2}}-m_{_{1}}\right)}\right)=a_{_{2G}}$$
The wedge only has a horizontal component of the acceleration, so plugging in $$a_{_{12}}$$ gives...
$$\vec{a}_{_{2G}}=-\left(\frac{g}{cos\theta\left(\frac{m_{_{1}}}{m_{_{2}}-m_{_{1}}}+1\right)cot\theta+sin\theta}\right)\left(\frac{m_{_{1}}cos\theta}{\left(m_{_{2}}-m_{_{1}}\right)}\right)\hat{i}$$

Note the minus sign; the block accelerates to the left, while our +i is to the right.

All that is left is finding the acceleration of 1 relative to the ground.

Recalling

$$\vec{a}_{_{1G}} = \vec{a}_{_{2G}} + \vec{a}_{_{12}}$$
$$\vec{a}_{_{1G}} = -a_{_{2G}}\hat{i} + a_{_{12X}}\hat{i}+a_{_{12Y}}\hat{j}$$
$$\vec{a}_{_{1G}} = -a_{_{2G}}\hat{i} + a_{_{12}}cos\theta\hat{i}+a_{_{12}}sin\theta\hat{j}$$
$$\vec{a}_{_{1G}} = -a_{_{2G}}\hat{i} + a_{_{12}}(cos\theta\hat{i}+sin\theta\hat{j})$$

We can finish this problem by plugging in values for $$a_{_{12}}$$ and $$a_{_{2G}}$$

Therefore,
$$\vec{a}_{_{1G}} = - \left(\frac{g}{cos\theta\left(\frac{m_{_{1}}}{m_{_{2}}-m_{_{1}}}+1\right)cot\theta+sin\theta}\right)\left(\frac{m_{_{1}}cos\theta}{\left(m_{_{2}}-m_{_{1}}\right)}\right)\hat{i} + \left(\frac{g}{cos\theta\left(\frac{m_{_{1}}}{m_{_{2}}-m_{_{1}}}+1\right)cot\theta+sin\theta}\right)(cos\theta\hat{i}+sin\theta\hat{j})$$

Cleaning it up we end with a still messy solution:
$$\vec{a}_{_{1G}} = \left(\frac{g}{cos\theta\left(\frac{m_{_{1}}}{m_{_{2}}-m_{_{1}}}+1\right)cot\theta+sin\theta}\right)\left(cos\theta\left(\frac{m_{_{1}}}{m_{_{2}}-m_{_{1}}}\right)\hat{i}+sin\theta\hat{j}\right)$$

-Ataman

2. May 3, 2008

### Shooting Star

This actually simplifies to:

$$- \frac{{m_1 g\cos \theta \sin \theta }}{{m_2 - m_1 \sin ^2 \theta }}.$$

Notice that the denominator can become zero for certain possible values, which shows that something is wrong.

The correct answer is:

$$- \frac{{m_1 g\cos \theta \sin \theta }}{{m_2 + m_1 \sin ^2 \theta }}.$$

If the $\left(\frac{m_{_{1}}}{m_{_{2}}-m_{_{1}}}+1\right)$ was replaced by $\left (1 - \frac{{m_1 }}{{m_1 + m_2 }}\right)$, then you would get the correct answer. Naturally, the other acceleration values will also change. Check for mistakes. Apart from this, a very good effort indeed.

Last edited: May 3, 2008
3. May 3, 2008

### Ataman

Many thanks for pointing those out, Shooting Star. All of my mistakes came from several minus signs I forgot to put in for the acceleration of 2 relative to the ground $$(a_{_{2GX}})$$.

Here is a more correct version, for those who care. :)

-Ataman

4. May 4, 2008

### reilly

The problem is much more simple than you have indicated, with all due respect. My clue: consider the role of momentum and energy conservation.
Regards,
Reilly Atkinson

5. May 5, 2008

### Shooting Star

Need for an shorter solution

Hi Ataman,

You must have understood, in spite of your very correct proof, that a simpler solution is called for. Reilly Atkinson has given you the correct idea. Why don't you give it a try? We are all here to help you if you get stuck.