August 1st, 2019, 12:25 PM  #1 
Senior Member Joined: Mar 2019 From: TTF area Posts: 106 Thanks: 1  weight
The diagram shows a viewing platform near a lake. The end of the platform juts out to the lake. Without the force F the platform is just about to tip. a. Use the data given in the diagram to calculate the force F required preventing the platform from tipping. Last edited by skipjack; August 2nd, 2019 at 02:20 PM. 
August 1st, 2019, 02:25 PM  #2  
Member Joined: Jun 2019 From: AZ, Seattle, San Diego Posts: 30 Thanks: 21  Quote:
I'm asking because sometimes you don't finish your threads and we find out later (after spending time to help you) that you really posted the exercise as practice for visitors (i.e., you're not interested in completing the exercise). Meanwhile, here's a beginning question. Do you understand the label 2500N?  
August 1st, 2019, 02:38 PM  #3 
Senior Member Joined: Mar 2019 From: TTF area Posts: 106 Thanks: 1 
I actually want to know how to solve it. 2500N is 2500Newtons.
Last edited by helpmeddddd; August 1st, 2019 at 02:44 PM. 
August 1st, 2019, 03:02 PM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587 
Rotational equilibrium about any point (you get to choose that pivot point) I assume you know how to calculate individual torques about a point. $\displaystyle \tau_{net} = 0 \implies \sum \tau_{ccw} = \sum \tau_{cw}$ I would use the point marked in red on the attached diagram ... 
August 1st, 2019, 03:20 PM  #5 
Senior Member Joined: Mar 2019 From: TTF area Posts: 106 Thanks: 1 
I actually have to do it for the vector F. So what would I need to do first?

August 1st, 2019, 05:01 PM  #6  
Math Team Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587  Quote:
force F exerts a counterclockwise torque the horizontal beam's center of mass exerts a counterclockwise torque the weight at the far right exerts a clockwise torque for rotational equilibrium ... sum of the two counterclockwise torques = the single clockwise torque set up an algebraic equation and solve for F note ... since all forces are acting perpendicular to the horizontal beam, torque = force x horizontal distance from the pivot  
August 1st, 2019, 05:33 PM  #7 
Senior Member Joined: Mar 2019 From: TTF area Posts: 106 Thanks: 1 
I'm not confident with is but is it, force x distance 2500x3=7500N

August 1st, 2019, 10:50 PM  #8  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,257 Thanks: 927 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Look, you've had two posts now that are telling you that you need to find the torque on all the forces, not just the one you are looking for. Why don't you listen to someone's advice this time and take a look at what skeeter wrote instead of trying to find a way to get us to do the problem for you. Dan  
August 2nd, 2019, 03:25 AM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,921 Thanks: 2203 
Are the platform's supports attached to the ground and therefore unable to move? Also, is neither support attached to the platform?

August 2nd, 2019, 12:54 PM  #10 
Senior Member Joined: Mar 2019 From: TTF area Posts: 106 Thanks: 1 
The torque on the 2500N (far right of the diagram) should be 2500Nm ( I used the 1m for distance in the equation) The torque for 1500N (IN the middle of diagram) I got 3000Nm(I used the 2m measurement) The torque for F is? ( I'd use 3m for distance) 

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