It wont let me add a photo for some reason. Do you know any alternatives?
Courtesy of notnek:
Some users experience problems with attachments if they upload the image direct to TSR. If you are having problems then we recommend uploading the image to Imgur instead:
1) Go to http://imgur.com/ then click New Post at the top then browse to find your image. 2) Once the image is displayed, right-click on the image and choose "Copy Image Location". It is important you do this so you get the direct image link as opposed to the Imgur link. 3) To display the image in a TSR post you can paste this copied link into your post surrounded by tags e.g. .
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 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?
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.....
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.....
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.