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| Both sides previous revision Previous revision Next revision | Previous revision | ||
| 183_notes:examples:walking_in_a_boat [2014/10/01 05:43] – pwirving | 183_notes:examples:walking_in_a_boat [2014/10/01 05:51] (current) – pwirving | ||
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| $x_{cm,f} = \dfrac {M(x+D+\dfrac{L}{2}) + m(x+D)}{M+m}$ | $x_{cm,f} = \dfrac {M(x+D+\dfrac{L}{2}) + m(x+D)}{M+m}$ | ||
| + | As indicated the center of the mass of the system as not changed position therefore: | ||
| $x_{cm,f} = x_{cm,i}$ | $x_{cm,f} = x_{cm,i}$ | ||
| + | |||
| + | So we can relate the equations for initial and final to each other: | ||
| $\dfrac {M(x+D+\dfrac{L}{2}) + m(x+D)}{M+m} = \dfrac {M(D+\dfrac{L}{2}) + m(D+L)}{M+m}$ | $\dfrac {M(x+D+\dfrac{L}{2}) + m(x+D)}{M+m} = \dfrac {M(D+\dfrac{L}{2}) + m(D+L)}{M+m}$ | ||
| - | Same denominator | + | Both of these equations have the same denominator |
| ${M(x+D+\dfrac{L}{2}) + m(x+D)} = {M(D+\dfrac{L}{2}) + m(D+L)}$ | ${M(x+D+\dfrac{L}{2}) + m(x+D)} = {M(D+\dfrac{L}{2}) + m(D+L)}$ | ||
| - | Solve for x, | + | Multiply out to solve for x: |
| - | $M_x + M(D + \dfrac{L}{2}) + mx + mD = M(D+\dfrac{L}{2}) + mD + mL$ | + | $Mx + M(D + \dfrac{L}{2}) + mx + mD = M(D+\dfrac{L}{2}) + mD + mL$ |
| - | Cancel like terms | + | Cancel like terms we get: |
| $Mx + mx = mL$ | $Mx + mx = mL$ | ||
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| $(M+m)x = mL$ | $(M+m)x = mL$ | ||
| + | |||
| + | Therefore: | ||
| $x = (\dfrac{m}{M+m})L$ | $x = (\dfrac{m}{M+m})L$ | ||
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| Does this make sense? | Does this make sense? | ||
| + | |||
| + | If you want to check whether something makes sense a good start is to check the units: | ||
| Units $(x) = m$ $(\dfrac{m}{M+m})$ = unitless | Units $(x) = m$ $(\dfrac{m}{M+m})$ = unitless | ||
| - | If M is really big then $x \equiv 0$, think oil thanker | + | Therefore m is the remaining unit. |
| + | |||
| + | Another check is that we know that if M is really big then $x \equiv 0$, think of a similar scenario to one just discussed occurring on a oil thanker. | ||
| $x = (\dfrac{m}{M+m})L | $x = (\dfrac{m}{M+m})L | ||
| - | If M = 0 then x $\equiv L$, no boat limit | + | Another check would be to check what happens when M = 0 then x $\equiv L$, i.e. if there was no boat. |
| $x = (\dfrac{m}{M+m})L | $x = (\dfrac{m}{M+m})L | ||
| So the motion of the center of mass of a system is dictated by the net external force. | So the motion of the center of mass of a system is dictated by the net external force. | ||