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August 1st, 2019, 12:25 PM   #1
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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.
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 Screenshot 2019-08-02 at 8.24.29 AM.png (5.6 KB, 7 views)

Last edited by skipjack; August 2nd, 2019 at 02:20 PM.

August 1st, 2019, 02:25 PM   #2
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Quote:
 Originally Posted by helpmeddddd … calculate the force F required [to prevent] the platform from tipping.
Hello. Did you start this thread because you're actually trying to solve part (a), or did you post it as practice for other people?

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
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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 ...
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 Equilibrium5.jpg (11.3 KB, 1 views)

 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
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Quote:
 Originally Posted by helpmeddddd I actually have to do it for the vector F. So what would I need to do first?
relative to the pivot ...

force F exerts a counter-clockwise torque

the horizontal beam's center of mass exerts a counter-clockwise torque

the weight at the far right exerts a clockwise torque

for rotational equilibrium ...

sum of the two counter-clockwise 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
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Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 Originally Posted by helpmeddddd I'm not confident with is but is it, force x distance 2500x3=7500N
Where did you get the 3 from?

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? Thanks from topsquark
 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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