183_notes:examples:holding_block_against_a_wall

Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Both sides previous revision Previous revision
Next revision		 Previous revision
183_notes:examples:holding_block_against_a_wall [2014/09/22 04:06] – pwirving183_notes:examples:holding_block_against_a_wall [2014/09/22 04:16] (current) – pwirving
Line 30: Line 30:
 ==== 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:Δpy=(FN−mg)Δt,assumingitslides y:Δpy=(FN−mg)Δt,assumingitslides
  
-Combining these two equations, we have+Combining these two equations 
 + 
 +(Fhand−FN)Δt=0  
 + 
 +$ F_{hand} \Delta t - F_N \Delta t = 0  \,\,\,\,\,\,\,Multiply\, out.$ 
 + 
 +FhandΔt=FNΔtMakeequaltoeachother. 
 + 
 +Fhand=FNCancelΔt. 
 + 
 +Substituting in we get:
  
 Δpy=(Fhead−mg)Δt=(0.6(40N)−(3kg)(9.8N/kg))Δt Δpy=(Fhead−mg)Δt=(0.6(40N)−(3kg)(9.8N/kg))Δt
  • 183_notes/examples/holding_block_against_a_wall.1411358807.txt.gz
  • Last modified: 2014/09/22 04:06
  • by pwirving