What is the angle between the wire and the vertical?

[Peter G.]

A stone weight 32 N is attached to a wire and hung from a rigid support. A string is then attached to the bob and is pulled sideways with a horizontal force P until the tension in the wire is 44N

a) What is the angle between the wire and the vertical?
b) Calculate the size of the force P
c) Repeat your calculations for the case when the tension is 140 N

a) cos (32/44) Angle = 43.34
b)44^2 - 32^2 = P ^2
P = 30.20 N

Same methods for c) 77.79 degrees and 136.29 N

Think those are correct?

For the next question, I had a lot of difficulty, so I dealt with the information I had but I don't know if what I did is right: A framed picture of weight 15 N is to be hung on the wall, using a piece of string. The ends of the string are tied to two points, 0.60m apart, on the same horizontal level, on the back of the picture. Draw a free body force diagram for the picture and find the tension in the string if: a) the string is 1.0 m long, b) the string is 0.66 m

So, since the string is tied on the same horizontal level, I think that the length of the string for the left and right sides will be the same, so, for a) 0.50 m and 0.50 m. With that and half of the 0.60 m, I formed a triangle, I managed to find the angle to the vertical. With the angle to the vertical and using the vertical component of tension, I assumed that it was 15/2 for each, thus, 7.5 m I got: T = 7.5/cos 36.86989765, which is = to: 9.38 N

I did the same for part B and got 18.0 N.

What I did is acceptable/correct?

Thanks in advance,
Peter G.

 

[Delphi51]

a) cos (32/44) Angle = 43.34

I don't understand this and got a slightly different answer.
I agree with the P = 30.2, though. Curious; I needed the angle to find the 30.2.

 

[Peter G.]

They want the angle to the vertical: We have the vertical component of the tension, which is 32 N and the tension, 44 N. So I did cos of x = Adjacent Side / Hypothenuse, hence, 32/44 and got that angle.

 

[Delphi51]

Okay, so you have cos(A) = P/44.
I wrote that the sum of the horizontal forces is zero (F = ma = 0) and got the same thing:
P - 44*cos(A) = 0. This has two unknowns, so can't do anything with it yet.

For the vertical forces, we have 44*sin(A) - 32 = 0.
This one can be solved for sin(A) = 32/44
A = inverseSin(32/44)

 

[Peter G.]

Delphi, I think you are finding the angle to the horizontal, because if you add the angle I got to yours + 90 degrees we get 180 degrees. But one way or the other, I understood what you did. But I think, since we needed the angle to the vertical it was a bit easier because we already had two sides of the triangle: The tension (hypothenuse) and the vertical component of the tension, which is equal to the weight of the stone, so I did inverseCos (32/44)

Do you think you could help me with the next question?

Thanks,
Peter G

 

[Delphi51]

Oh, yes, I found the angle with horizontal. Sorry!

I agree with your answer for 2a.

 

[Peter G.]

No problem

I used the same method for 2 B, so I assume it is right too, as long as I did not make any calculations mistake!

Thanks Delphi

 

[Peter G.]

Oh, just another thing. The question (2 a and b) asks for the tension in the string and it says that one string is used to hold the picture. So: The answer would be, for a 2 x (7.5 / cos 36.86989765) ?

Thanks,
Peter G