Fig 2 (below) shows a vase placed on a uniform shelf that is supported by a steel wire (orange line below). The ma ss of the vase is 0.65kg and the mass of the shelf is 2.0kg. The shelf is hinged at A. The steel wire is attached to the shelf 0.30m from A and is at an angle of 30 degrees to the shelf. The other end of the steel wire is attached to the wall.
Show, by taking moments about A, that the tension in the steel wire is about 50 N
So I thought I would take moments about A as the question asked and I did
A = (9,81*0.65*0.45) + (2*9.81*0.25) [0.25 because uniform beam so all mass acts here]
Then I thought I would get the force on A and then do A/sin(30). I got a value that was no where near 50N. Anyone have any ideas? Thanks.
Your working is correct. But let's not forget that the string is anchored to the shelf 0.3m away. Therefore it is not sin30A, it is the counterclockwise moment of 0.3sin30A Which will yield 51N
Your working is correct. But let's not forget that the string is anchored to the shelf 0.3m away. Therefore it is not sin30A, it is the counterclockwise moment of 0.3sin30A Which will yield 51N
Thanks for the reply
I sort of understand why you did *0.3 but I don't know why you did it to the sin(30)? I thought the force acting on A would just be trig and not involve the 0.3? I'm still a bit confused! Sorry if I sound stupid! I just suck at the mech part of phys
I sort of understand why you did *0.3 but I don't know why you did it to the sin(30)? I thought the force acting on A would just be trig and not involve the 0.3? I'm still a bit confused! Sorry if I sound stupid! I just suck at the mech part of phys
Here, ill draw a diagram for you just a sec, ill edit this in a min. And no you dont sound stupid at all! its not easy stuff!
This looks like a question from one of the Physics specimen papers. Nice
Yup, it is. Just got round to finishing all the questions but I had to go back to this one because I couldn't get anywhere near 50N! Turns out it was that one silly step with the *0.3
You need to work out the vertical force upwards from the rope A(7.78)=0.3*? So ?=25.9 Then just use trigonometry to find the force along the wire which is at 60 to the vertical