I took moments about X and as the system is in equilibrium, I've said that 6gcos10+2Tycos55cos80=2Tysin55cos which leads to Ty=1.457.9=41.4N which is incorrect. As this is incorrect, my other results would also be wrong so I'm stuck.
I took moments about X and as the system is in equilibrium, I've said that 6gcos10+2Tycos55cos80=2Tysin55cos which leads to Ty=1.457.9=41.4N which is incorrect. As this is incorrect, my other results would also be wrong so I'm stuck.
Is my model wrong?
Should the answer be 40.9? If so, all that is wrong is that you have rounded your numbers off before you got to the final answer.
Thanks for replying . The answers are Ty=37.4N, Tx=31.9N and 29.3∘ to the vertical.
I asked a friend who approached the problem differently by considering a converging triangle of forces.
Working
I follow it but I'm unsure what it is I did wrong. Can you help?
I also did it with a triangle of forces (effectively) and got the same answer as you, give or take the rounding issue.
The difference between your friend's version and ours is the direction the strings are going in. You and I both had them going outwards from the ends of the rod, he has them coming inwards.
A careful reading of the question says that the string from Y is fixed to the ceiling at a point above the rod. I had just taken that as meaning that it was fixed higher than the rod (which is .... obvious) but it actually means vertically above some point on the rod, which is why your friend's diagram is correct.
I also did it with a triangle of forces (effectively) and got the same answer as you, give or take the rounding issue.
The difference between your friend's version and ours is the direction the strings are going in. You and I both had them going outwards from the ends of the rod, he has them coming inwards.
A careful reading of the question says that the string from Y is fixed to the ceiling at a point above the rod. I had just taken that as meaning that it was fixed higher than the rod (which is .... obvious) but it actually means vertically above some point on the rod, which is why your friend's diagram is correct.
I repeated the question drawing the forces coming inwards and I still got the same answer I'm afraid. I honestly don't get what's wrong with what I did (beyond the rounding error). What's the significance of "vertically above the rod" sorry? I'm sorry to be a bit of a nuisance here but it's a bit frustrating not knowing what went wrong. I wouldn't have an answer in the exam obviously.
You should also practice resolving distances rather than forces - I find this makes certain scenarios easier to deal with.
Thanks. I think that was the point of the exercise .
Do you know why it doesn't work out for if I use 6g as part of the triangle? Like in the example? What's different about the example?
Would you mind assisting @ghostwalker please? When I used the triangle my friend set up, I was able to follow it to get to the correct answer but I'm uncomfortable not knowing what went wrong with my working. Even when I drew my lines going inwards, I got the same figures and answer.
Hi, there sorry. The original question is reposted below. I redrew the diagram with the forces pointed inwards and taking moments about X I got 6gcos10+2Tysin35sin10=2Tycos35cos10⇒Ty=1.41357.9=40.98. There may be a rounding error but still, it's identical to what I've below and far from the actual answer. I'm unsure why it doesn't work.
Tiny Hobbit has said A careful reading of the question says that the string from Y is fixed to the ceiling at a point above the rod. I had just taken that as meaning that it was fixed higher than the rod (which is .... obvious) but it actually means vertically above some point on the rod, which is why your friend's diagram is correct.
But the significance of what she says is lost on me. Does it affect my model?
Hi, there sorry. The original question is reposted below. I redrew the diagram with the forces pointed inwards and taking moments about X I got 5gcos10+2Tysin35sin10=2Tycos35cos10⇒Ty=1.41357.9=40.98. There may be a rounding error but still, it's identical to what I've below and far from the actual answer. I'm unsure why it doesn't work.
Your rod seems to have become lighter. 5kg rather than the 6kg in the question. Your Tysin35sin10 should be going anticlockwise and as such on the RHS of your equation (with the 2, of course).
Tiny Hobbit has said A careful reading of the question says that the string from Y is fixed to the ceiling at a point above the rod. I had just taken that as meaning that it was fixed higher than the rod (which is .... obvious) but it actually means vertically above some point on the rod, which is why your friend's diagram is correct.
But the significance of what she says is lost on me. Does it affect my model
Yes - your model (final image) is of the orignal set up.
Your rod seems to have become lighter. 5kg rather than the 6kg in the question. Your Tysin35sin10 should be going anticlockwise and as such on the RHS of your equation (with the 2, of course).
Yes - your model (final image) is of the orignal set up.
Thanks ghostwalker. The 6 was a typo on here and I suppose it makes sense for the split components of the Ty to be both upwards if it was upwards in the first place.
Thanks again for your help. I got 31.95 which is not quite 31.9 (another rounding error) which suggests I should use more d.p in prior steps than I do or keep the trigfunction(value) until the last step.
I suppose it makes sense for the split components of the Ty to be both upwards if it was upwards in the first place.
That sounds as if you're a bit unsure. The Ty sin 35 is above the line of the rod, so if you want the component perpendicular to the rod, it will be above, hence anticlockwise.
Thanks again for your help. I got 31.95 which is not quite 31.9 (another rounding error) which suggests I should use more d.p in prior steps than I do or keep the trigfunction(value) until the last step.
As a general guide, I delay putting in actual numbers as long as possible - you often find the algebra will simplify.
Regarding the question in general: If you'd done a bit of elmentary geometry at the start, and resolved Ty parallel and perpenedicular to the rod (rather than going for vertical and horizontal as an intermediate resolution), the algebra would have been a lot simpler.
That sounds as if you're a bit unsure. The Ty sin 35 is above the line of the rod, so if you want the component perpendicular to the rod, it will be above, hence anticlockwise.
Thank
If you'd done a bit of elmentary geometry at the start, and resolved Ty parallel and perpenedicular to the rod (rather than going for vertical and horizontal as an intermediate resolution), the algebra would have been a lot simpler.
I'll type my next question this time as I know that pictures aren't always very clear
But it's question 8 in exercise 5D of the Pearsons Edexcel book if you have it to hand
Question
A uniform ladder rests in limiting equilibrium with one end on rough horizontal ground and the other end against a rough vertical wall. The coefficient of friction between the ladder and the ground is μ1 and the coefficient of friction between the ladder and the wall is μ2. Given the ladder makes and angle θ with the horizontal, show that tanθ=2μ11−μ1μ2
μ1=RAF1μ2=RBF2 tanθ=2μ11−μ1μ2=RA2F11−RAF1RBF2=2F1RbF1(RA−F2)=2RBRa−F2=tanθ as required. So my question is that if I've shown tanθ=... and I show that something else is =..., is that fine? It seemed simpler than changing ... into the LHS of the question.
μ1=RAF1μ2=RBF2 tanθ=2μ11−μ1μ2=RA2F11−RAF1RBF2=2F1RbF1(RA−F2)=2RBRa−F2=tanθ as required. So my question is that if I've shown tanθ=... and I show that something else is =..., is that fine? It seemed simpler than changing ... into the LHS of the question.
I'm not quite sure what you mean. Do you mean leaving it as the tantheta =forces part?