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| Both sides previous revision Previous revision Next revision | Previous revision | ||
| 183_notes:examples:sledding [2014/10/13 05:15] – pwirving | 183_notes:examples:sledding [2014/10/22 04:06] (current) – pwirving | ||
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| - | ===== Example: | + | ===== Example: |
| A little girl is riding her sled on a hill. If she starts a distance d up the hill, which makes an angle θ with the horizontal, how far will she travel along the flat snowy ground? | A little girl is riding her sled on a hill. If she starts a distance d up the hill, which makes an angle θ with the horizontal, how far will she travel along the flat snowy ground? | ||
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| {{course_planning: | {{course_planning: | ||
| - | Need to find f_{1} & f_{2} | + | Need to find $f_{1}$ & $f_{2}$ |
| To find F1 we can say that the sum of the forces in the x direction are equal to ma1 But we don't need this because we know that f1=μkN. | To find F1 we can say that the sum of the forces in the x direction are equal to ma1 But we don't need this because we know that f1=μkN. | ||
| - | $\sum{F_{x}} = f_{1} - mgsinθ = ma_{1} | + | $\sum{F_{x}} = f_{1} - mgsinθ = ma_{1}$ |
| The sum of the forces in the y direction we do need because this allows us to express N. | The sum of the forces in the y direction we do need because this allows us to express N. | ||
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| ΔUg=−(μkmgcosθ)d−(μkmg)x | ΔUg=−(μkmgcosθ)d−(μkmg)x | ||
| + | |||
| + | Substitute in the equation for gravitational potential energy for ΔUg | ||
| +mg(yf−yi)=−μkmgdcosθ−μkmgx | +mg(yf−yi)=−μkmgdcosθ−μkmgx | ||
| + | |||
| + | Rearrange to get the following expression. | ||
| yf−yi=−μk(dcosθ+x) | yf−yi=−μk(dcosθ+x) | ||
| - | What is yf−yi in terms of what we know? | + | What is yf−yi in terms of what we know? Eventually we want to express x in terms of variables we know. |
| {{course_planning: | {{course_planning: | ||
| + | |||
| + | From the diagram of the incline we get: | ||
| yf−yi=−dsinθ | yf−yi=−dsinθ | ||
| + | |||
| + | Substitue −dsinθ for yf−yi and then rearrange to express x in terms of known variables. | ||
| −dsinθ=−μk(dcosθ+x) | −dsinθ=−μk(dcosθ+x) | ||
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| x=d(sinθ−μkcosθμk) | x=d(sinθ−μkcosθμk) | ||
| + | |||
| + | A check of the units reveals that: | ||
| [x]=m | [x]=m | ||
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| [d]=m | [d]=m | ||
| - | All other quantities are unitless. | + | Which makes sense as all the other quantities are unit less. |
| + | E=γmc2 | ||