Differences
This shows you the differences between two versions of the page.
Next revision | Previous revision | ||
183_notes:examples:holding_block_against_a_wall [2014/09/16 06:52] – created pwirving | 183_notes:examples:holding_block_against_a_wall [2014/09/22 04:16] (current) – pwirving | ||
---|---|---|---|
Line 6: | Line 6: | ||
=== Facts ==== | === Facts ==== | ||
+ | The metal block has a mass of 3 kg | ||
+ | |||
+ | Horizontal force applied to metal block of 40N in positive x-direction | ||
+ | |||
+ | Coefficient of friction for the metal-wall pair of materials is 0.6 for both static and sliding friction. | ||
=== Lacking === | === Lacking === | ||
+ | $\vec{F}_{net}$ in the x-direction | ||
+ | |||
+ | $\vec{F}_{net}$ in the y-direction | ||
=== Approximations & Assumptions === | === Approximations & Assumptions === | ||
+ | Assume applied horizontal force is constant. | ||
=== Representations === | === Representations === | ||
+ | {{183_notes: | ||
+ | |||
+ | $\Delta \vec{p} = \vec{F}_{net} \Delta t$ | ||
==== 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 | + | $ x: \Delta p_x = (F_{hand} |
+ | |||
+ | $ y: \Delta p_y = (F_N - mg) \Delta t, \, | ||
+ | |||
+ | 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 = (-5.4 N) \Delta t $ | ||
+ | |||
+ | Since there is a nonzero impulse in the -y direction, the block will slip downward with increasing speed. | ||
+ | |||