Physics Help Forum Pressure acting on wedge
 User Name Remember Me? Password

 Advanced Mechanics Advanced Mechanics Physics Help Forum

 Jan 15th 2016, 05:02 AM #1 Senior Member   Join Date: Jun 2014 Posts: 306 Pressure acting on wedge 1.) for the Sum of Fx , the author gave P1Δx - P3 (I sin theta) , why the author left off the sin theta at the below there ? 2.) Why the author assume Δz = 0 ? What's the purpose of assuming Δz = 0 ? image here: http://imgur.com/9M19HN9 http://imgur.com/0gMXWHS
Jan 15th 2016, 07:11 AM   #2
Physics Team

Join Date: Jun 2010
Location: Morristown, NJ USA
Posts: 2,320
 Originally Posted by ling233 1.) for the Sum of Fx , the author gave P1Δx - P3 (I sin theta) , why the author left off the sin theta at the below there ?
Because it cancels out. Starting with equation (3-3a) sub in delta_x = l sin(theta) and you get:

P_1 l sin(theta) - P_3 l sin(theta) = 0, so:
P_1 - P_3 = 0.

 Originally Posted by ling233 2.) Why the author assume Δz = 0 ? What's the purpose of assuming Δz = 0 ?
I must admit the author doesn't explain this very well, but it seems his purpose is to show that the pressure at a point (i.e. where delta_z = delta_x = delta_y = 0) is the same in all directions. So why doesn't he state that delta_x and delta_y also go to zero? Because those quantities have already cancelled out of the equations, so he ignores them.

Jan 15th 2016, 07:52 AM   #3
Senior Member

Join Date: Jun 2014
Posts: 306
 Originally Posted by ChipB Because it cancels out. Starting with equation (3-3a) sub in delta_x = l sin(theta) and you get: P_1 l sin(theta) - P_3 l sin(theta) = 0, so: P_1 - P_3 = 0. I must admit the author doesn't explain this very well, but it seems his purpose is to show that the pressure at a point (i.e. where delta_z = delta_x = delta_y = 0) is the same in all directions. So why doesn't he state that delta_x and delta_y also go to zero? Because those quantities have already cancelled out of the equations, so he ignores them.
what do you mean by Because those quantities have already cancelled out of the equations, so he ignores them??

Jan 15th 2016, 10:14 AM   #4
Physics Team

Join Date: Jun 2010
Location: Morristown, NJ USA
Posts: 2,320
 Originally Posted by ling233 what do you mean by Because those quantities have already cancelled out of the equations, so he ignores them??
Do the math! Starting with equation (3-3b), replace the "l cos(theta)" term with delta_x, then divide through by delta_x.

Jan 15th 2016, 05:26 PM   #5
Senior Member

Join Date: Jun 2014
Posts: 306
 Originally Posted by ChipB Because it cancels out. Starting with equation (3-3a) sub in delta_x = l sin(theta) and you get: P_1 l sin(theta) - P_3 l sin(theta) = 0, so: P_1 - P_3 = 0. I must admit the author doesn't explain this very well, but it seems his purpose is to show that the pressure at a point (i.e. where delta_z = delta_x = delta_y = 0) is the same in all directions. So why doesn't he state that delta_x and delta_y also go to zero? Because those quantities have already cancelled out of the equations, so he ignores them.
how about pressure at othert point other than delta_z = delta_x = delta_y = 0 ? are the pressure same at all direction ?

 Jan 15th 2016, 06:49 PM #6 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,320 Your question makes no sense. By definition a point has zero dimension, so delta-x = delta-y = delta_x = 0 for ALL points.

 Tags acting, pressure, wedge

 Thread Tools Display Modes Linear Mode

 Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post ling233 Advanced Mechanics 2 Feb 26th 2016 09:44 AM Luke Advanced Mechanics 4 Dec 2nd 2013 05:29 PM SMA777 Kinematics and Dynamics 3 Jan 12th 2012 09:28 AM caffe Kinematics and Dynamics 3 Jan 1st 2012 08:13 AM cyt91 Light and Optics 3 Jun 27th 2010 07:17 PM