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Further maths forces help

Photo is below Please do then explain the process for me.

Thank you
I can't see a photo.
Reply 2
Original post by old_engineer
I can't see a photo.


It wont let me add a photo for some reason. Do you know any alternatives?
Original post by amin11234
It wont let me add a photo for some reason. Do you know any alternatives?


Courtesy of notnek:

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Reply 4
Original post by old_engineer
I can't see a photo.


Here it is sorry

https://imgur.com/gallery/Riwkm
Original post by amin11234


OK the suggested approach to part (i) of this question is:

1) Take moments about the bottom of the ladder, noting that the force exerted by the wall on the top of the ladder is purely horizontal. (It can’t have a vertical component because the wall is smooth).

2) Resolve forces horizontally. The ladder is in equilibrium, so the horizontal reaction at the wall must exactly balance the frictional force at the foot of the ladder.
Reply 6
Original post by old_engineer
OK the suggested approach to part (i) of this question is:

1) Take moments about the bottom of the ladder, noting that the force exerted by the wall on the top of the ladder is purely horizontal. (It can’t have a vertical component because the wall is smooth).

2) Resolve forces horizontally. The ladder is in equilibrium, so the horizontal reaction at the wall must exactly balance the frictional force at the foot of the ladder.


Ok thanks but how do you show that F=90(1+x)tan20?
Original post by amin11234
Ok thanks but how do you show that F=90(1+x)tan20?


If you take moments about the bottom of the ladder as suggested, everything should become clear. If it doesn't become clear, please post your working.
Reply 8
Original post by amin11234
Photo is below Please do then explain the process for me.

Thank you


https://imgur.com/gallery/Riwkm


This is the same image you posted three hours ago.
Reply 10
Original post by old_engineer
If you take moments about the bottom of the ladder as suggested, everything should become clear. If it doesn't become clear, please post your working.


https://imgur.com/gallery/MOKkw


The diagram is very difficult to read, but from what I can see, it may be useful for you to note that cos 70 is the same as sin 20. If you make that change you will have two sin 20 terms on one side of the equation and a cos 20 term on the other side, and you are aiming for an expression that involves tan 20.....
Reply 12
Original post by old_engineer
The diagram is very difficult to read, but from what I can see, it may be useful for you to note that cos 70 is the same as sin 20. If you make that change you will have two sin 20 terms on one side of the equation and a cos 20 term on the other side, and you are aiming for an expression that involves tan 20.....


ok, thank you for your help, finally got it.
Reply 13
Original post by old_engineer
This is the same image you posted three hours ago.


forces topic assessment q3.JPG

One more thing, for this question would there be two reaction forces: one at B and C or would there just be one on C?

thanks
Original post by amin11234


One more thing, for this question would there be two reaction forces: one at B and C or would there just be one on C?

thanks


You should reckon on there being reaction forces at both B and C. Bear in mind, though, that the reaction at C could be zero if the bridge was on the point of rotating anticlockwise about B. This would typically come about through the addition of an additional mass somewhere on the bridge.

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