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183_notes:examples:holding_block_against_a_wall [2014/09/22 04:06] – pwirving | 183_notes:examples:holding_block_against_a_wall [2014/09/22 04:16] (current) – pwirving | ||
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==== Solution ==== | ==== Solution ==== | ||
- | You need to identify whether the momentum in the y direction is negative (if it is, that would mean it was slipping down the wall). | + | You need to identify whether the momentum in the y direction is negative (if it is, that would mean it was slipping down the wall, if it was positive it would mean it is slipping up the wall). |
Start by computing the change in momentum for both the x direction and the y direction. | Start by computing the change in momentum for both the x direction and the y direction. | ||
- | $ x: \Delta p_x = (F_{head} - F_N) \Delta t = 0 $ | + | $ x: \Delta p_x = (F_{hand} - F_N) \Delta t = 0 $ |
$ y: \Delta p_y = (F_N - mg) \Delta t, \, | $ y: \Delta p_y = (F_N - mg) \Delta t, \, | ||
- | Combining these two equations, we have | + | Combining these two equations |
+ | |||
+ | $ (F_{hand} - F_N) \Delta t = 0 $ | ||
+ | |||
+ | $ F_{hand} \Delta t - F_N \Delta t = 0 \,\, | ||
+ | |||
+ | $ F_{hand} \Delta t = F_N \Delta t \, | ||
+ | |||
+ | $ F_{hand} = F_N \, | ||
+ | |||
+ | Substituting in we get: | ||
$ \Delta p_y = (F_{head} - mg) \Delta t = (0.6(40N) - (3 kg)(9.8 N/kg)) \Delta t $ | $ \Delta p_y = (F_{head} - mg) \Delta t = (0.6(40N) - (3 kg)(9.8 N/kg)) \Delta t $ |