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183_notes:examples:holding_block_against_a_wall [2014/09/16 07:08] – pwirving | 183_notes:examples:holding_block_against_a_wall [2014/09/22 04:16] (current) – pwirving |
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=== Facts ==== | === Facts ==== |
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| The metal block has a mass of 3 kg |
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| Horizontal force applied to metal block of 40N in positive x-direction |
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| Coefficient of friction for the metal-wall pair of materials is 0.6 for both static and sliding friction. |
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=== Lacking === | === Lacking === |
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| →Fnet in the x-direction |
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| →Fnet in the y-direction |
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=== Approximations & Assumptions === | === Approximations & Assumptions === |
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| Assume applied horizontal force is constant. |
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=== Representations === | === Representations === |
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| {{183_notes:block_on_wall.jpg|}} |
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| Δ→p=→FnetΔt |
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==== Solution ==== | ==== Solution ==== |
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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). |
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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. |
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$ x: \Delta p_x = (F_{head} - F_N) \Delta t = 0 $ | $ x: \Delta p_x = (F_{hand} - F_N) \Delta t = 0 $ |
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y:Δpy=(FN−mg)Δt,assumingitslides | y:Δpy=(FN−mg)Δt,assumingitslides |
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Combining these two equations, we have | Combining these two equations |
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| (Fhand−FN)Δt=0 |
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| $ F_{hand} \Delta t - F_N \Delta t = 0 \,\,\,\,\,\,\,Multiply\, out.$ |
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| FhandΔt=FNΔtMakeequaltoeachother. |
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| Fhand=FNCancelΔt. |
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| Substituting in we get: |
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Δpy=(Fhead−mg)Δt=(0.6(40N)−(3kg)(9.8N/kg))Δt | Δpy=(Fhead−mg)Δt=(0.6(40N)−(3kg)(9.8N/kg))Δt |