Help on Mechanics Question: Find BC Angle with Vertical

[Gughanath]

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.

 

[HallsofIvy]

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 θ. 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(θ) (notice the negative- the two string pull in opposite directions horizontally) and the y component is 2Tcos(θ) (positive: both string pull upwards).
Since the object is not moving horizontally, the two x components must "offset":
T sin(30)- 2T sin(θ)= 0 or T sin(30)= 2T sin(θ).
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 θ?

 

[Gughanath]

I worked out the value of theta to be 90.4 degrees. I don't know where I am going wrong!

 

[Gughanath]

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.

 

[Gughanath]

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

 

[Gughanath]

The introductio of sin into the equation doesn't let me continue any further?

 

[Nylex]

Post your work, so we can help more!

 

[Doc Al]

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

 

[Gughanath]

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]

The second equation is all you need.

 

[Gughanath]

how? that doesn't 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]

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

Use your calculator.

 

[Gughanath]

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

Use your calculator.

Oh no! Did I just make the mistake of the century? This is soo embarassing. OBVIOuSLY 2T/T is not T!...SHOOT ME...and yes, what you said is right. How could I forget these simple rules? all right..let me try again

 

[Gughanath]

theta = 14.5 degrees!

 

[Doc Al]

You got it.