# A level maths mechanics ladders question

the first part of this question requires taking moments around A but when doing the force around B they used Nsintheta but wouldn’t the perpendicular force be Ncostheta
Original post by esha06
the first part of this question requires taking moments around A but when doing the force around B they used Nsintheta but wouldn’t the perpendicular force be Ncostheta
this is the only way i can share the pic
https://share.icloud.com/photos/0faNtthzD4VCSMvxelRZ12GpQ
Original post by esha06
this is the only way i can share the pic
https://share.icloud.com/photos/0faNtthzD4VCSMvxelRZ12GpQ

Not too sure what you mean. For part a) youd take moments about A (and trivially resolve vertically/horizontally). The perpendicular distance of N from A is effectively the height of the wall so 2asin(theta) and the ladders com is acos(theta) perpendicular distance.
Original post by mqb2766
Not too sure what you mean. For part a) youd take moments about A (and trivially resolve vertically/horizontally). The perpendicular distance of N from A is effectively the height of the wall so 2asin(theta) and the ladders com is acos(theta) perpendicular distance.

ignore my absolutely hideous diagram but what i’m tryna say is wouldn’t the moment around it be 2a x cos theta bc of the way u break up the force 2a is the distance and perpendicular to that is cos theta
https://share.icloud.com/photos/074r6EgnWjQv6uOobaAQ2wiQA
(edited 2 months ago)
Original post by esha06
ignore my absolutely hideous diagram but what i’m tryna say is wouldn’t the moment around it be 2a x cos theta bc of the way u break up the force
https://share.icloud.com/photos/074r6EgnWjQv6uOobaAQ2wiQA

No, the angle between the ladder and N is theta (alternate angles). The complementary angle between the ladder and the wall is 90-theta so your new angle is also 90-theta.
Original post by mqb2766
No, the angle between the ladder and N is theta (alternate angles). The complementary angle between the ladder and the wall is 90-theta so your new angle is also 90-theta.
so then the opposite angle would be theta like this?
https://share.icloud.com/photos/08eZPUtYi3SOiVByYD2OqjEBg
Original post by esha06
so then the opposite angle would be theta like this?
https://share.icloud.com/photos/08eZPUtYi3SOiVByYD2OqjEBg

Yes, your dashed line is parallel to the ladder the horizontal N line is parallel to the floor, so that angle must be theta.

Note that when youre working out moments you can think of either "perpendicular force" or "perpendicular distance". The latter (the height of the wall,, up to the ladder top) is arguably simpler here, so 2asin(theta)*force. Or even just use the definition of the moment as you know the angle, theta, between the ladder and N.
(edited 2 months ago)
Original post by mqb2766
Yes, your dashed line is parallel to the ladder the horizontal N line is parallel to the floor, so that angle must be theta.
Note that when youre working out moments you can think of either "perpendicular force" or "perpendicular distance". The latter (the height of the wall,, up to the ladder top) is arguably simpler here, so 2asin(theta)*force
thank youuuuu
Original post by esha06
thank youuuuu

A simple observation would be that W and N are perpendicular forces. Their corresponding moments must contain complementary trig terms.
Original post by mqb2766
A simple observation would be that W and N are perpendicular forces. Their corresponding moments must contain complementary trig terms.

oh that’s so smart i never would’ve thought about it like that 😭
Original post by esha06
oh that’s so smart i never would’ve thought about it like that 😭

NP. If you think of the perpendicular distances (rather than resolving forces so perpendicular forces), the weight corresponds to the floor (well half) and the normral reaction is the wall, so its easy to see the cos and sin respectively.