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Physics Help

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gnarrr
Posts: 823 - Karma: 12
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Please help me. I can't figure it out
Feb 23 2009 12:23AM
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flippintricker
Posts: 2408 - Karma: 9

it's been a while man. Just figure out weight of the box thing at the end, then use torques to get force on the cable?

Feb 23 2009 12:34AM
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gnarrr
Posts: 823 - Karma: 12
i already know the weight of the box is 9.8x85kg=833N but once i get that i have no idea where to go. i try to do something but i get the completely wrong answer.
Feb 23 2009 12:58AM
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slevin_kelevra
Posts: 416 - Karma: 13
find the force on the bar where the cable attaches.  The sum of forces in the y direction is equal to 0.  So it should be the opposite of the force(torque) where the beam is attached to the cable
Feb 23 2009 1:13AM
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spqr
Posts: 182 - Karma: 141
You should use torque equations to solve this. Torque is the length of the beam multiplied by force, multiplied by the sine of the angle the force makes with the moment arm. (6x833*sin75)

Then you need to figure out what force from the cable, will provide the necisary torque to keep things in equalibrium.

Summary: The sum of the torques and forces in equalibrium is zero. Find the torque from the mass, then use this to find what the tension is.
Feb 23 2009 1:32AM
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gnarrr
Posts: 823 - Karma: 12
i think i got it. i get 1511.6N as an answer.
Feb 23 2009 2:02AM
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B.G
Posts: 156 - Karma: 14
holy fuck my teach did that exact Q like a month ago

but i wasnt paying attention so i dont remeber

but use torques and remember torques only work when perpendicular to the board so itl be like mgcos(theta) or mgsin(theta) then just work it out around point p the atachment to the wall. then find the torque the wall needs to exert
Feb 23 2009 2:44AM
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B.G
Posts: 156 - Karma: 14
cable*
Feb 23 2009 2:44AM
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NESblowjob
Posts: 1174 - Karma: 41
try 6/4  X   cos 15    i think that might get you the answer let me know
Feb 23 2009 3:22AM
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Billek
Posts: 5665 - Karma: 98
Make sure to include the beam!
Feb 23 2009 7:04AM
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Billek
Posts: 5665 - Karma: 98
Nevermind!
Feb 23 2009 7:05AM
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