# Urgent help on mechanics!

I am revising for a test which is next week. I came across this question and I keep getting the wrong answer. Someone please help!

A particle has mass 2kg. It is attached at B to the end of two light inextensible strings AB and BC. When the particle is in equilibirum, AB makes an angle of 30 degrees with the vertical. The magnitude of the tension in BC is twice the magnitude of the tension in AB.

Find, in degress to one decimal place, the size of the angle that BC makes with the vertical.

## Answers and Replies

HallsofIvy
Homework Helper
Do this in terms of "components". Let T be the tension is the string AB. Then the x component of force for the AB string is Tcos(30) and the y component is Tsin(30).
We don't know the angle the BC string makes with the vertical so call it &theta;. We do know that the tension in the BC string is "is twice the magnitude of the tension in AB." so it is 2T. The x component of force for the BC string is -2T sin(&theta;) (notice the negative- the two string pull in opposite directions horizontally) and the y component is 2Tcos(&theta;) (positive: both string pull upwards).
Since the object is not moving horizontally, the two x components must "offset":
T sin(30)- 2T sin(&theta;)= 0 or T sin(30)= 2T sin(&theta;).
Since the object is not moving vertically, the two y components, together, must equal the weight of the object: Tcos(30)+ 2Tcos(&theta)= 2g.
Can you solve those two equations for T and &theta;?

I worked out the value of theta to be 90.4 degrees. I dont know where I am going wrong!

I think you did a lilttle mistake. The X component of the AB string should be Tcos 60 and the y components is Tcos 30.

oh sorry.my bad. Well I try again and get back to you!. Thanx

The introductio of sin into the equation doesnt let me continue any further???

Post your work, so we can help more!

Doc Al
Mentor
Hint: All you need consider are the horizontal components of the forces on the mass. Start there and show your work

ok. I got two equations. 2Tcosθ + Tcos30 = 2g............1
Tcos60 = 2Tcos(90-θ) ............2
I have two simultanous equations. I rearrange equation 2 to make T the subject, and then sub T into equation 1.

Am I going the right way?

Doc Al
Mentor
The second equation is all you need.

how? that doesnt make sense.
Tcos60 = 2Tcos (90-θ)
since cos 90 = 0, the equation simplifies to: Tcos60 = -2Tcos θ
cancel one T on each side>>>> cos60 = -tcosθ Now??

Doc Al
Mentor
FYI: In general, $\cos (A - B) \ne \cos A - \cos B$ !!!

Instead, cancel the T and get:
cos60 = 2cos (90-θ), and then
cos (90-θ) = (cos60)/2

Doc Al said:
FYI: In general, $\cos (A - B) \ne \cos A - \cos B$ !!!

Instead, cancel the T and get:
cos60 = 2cos (90-θ), and then
cos (90-θ) = (cos60)/2