Can someone check if I'm using the right method (forces question, AQA AS unit 2)

An object of weight 7.5N hangs on the end of a cord, which is attached to the midpoint of a wire stretched between two points on the same horizontal level, as shown in the picture below. Each half of the wire is at 12 degrees to the horizontal. Calculate the tension in each half of the wire.


Heres what I did:

Tension 1 = Tension 2

7.5N divided by 2 = 3.75N

so 3.75N is the vertical component of Tension 1 (T)

3.75 = Tsin12

T = 3.75/sin12

is this how most people would tackle this question? because I'm under the assumption that in general you shouldn't have to resort to dividing by 'sinx' (or cosx, tanx)

so is there a different method, or is this the standard way to do this question?

(if you have the nelson thornes AQA AS textbook, its question 4 on page 108)

"because I'm under the assumption that in general you shouldn't have to resort to dividing by 'sinx' (or cosx, tanx)"

It's the right idea.

You should balance the vertical forces.
Downwards you have the weight of the object and upwards you have the equal tensions in the two halves of the string.

The vertical components of those tensions are

T sin 12 + T sin 12
(or T cos 78 + T cos 78)

The equation you should write down is
7.5 = T sin 12 + T sin 12

this gives your answer.


There is no basis for your assumption



You often have to divide by sin, cos or tan. Why not?

An object of weight 7.5N hangs on the end of a cord, which is attached to the midpoint of a wire stretched between two points on the same horizontal level, as shown in the picture below. Each half of the wire is at 12 degrees to the horizontal. Calculate the tension in each half of the wire.




Heres what I did:

Tension 1 = Tension 2

7.5N divided by 2 = 3.75N

so 3.75N is the vertical component of Tension 1 (T)

3.75 = Tsin12

T = 3.75/sin12

is this how most people would tackle this question? because I'm under the assumption that in general you shouldn't have to resort to dividing by 'sinx' (or cosx, tanx)

so is there a different method, or is this the standard way to do this question?

(if you have the nelson thornes AQA AS textbook, its question 4 on page 108)

isnt it T cos (12), not T cos (78) as it's the angle made with th horizontal.....

No. It's the vertical components that balance the 7.5N weight.

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