resolving vertically - mechanics

Two fixed points A and B are on the same horizontal ceiling where AB = 24cm. A light elastic string has one end attached to A and the other end attached to B. The natural length of the string is 20cm. A particle of mass 0.4kg is attached to the mid-point of the string and hangs in equilibrium at a point C, which is 3.5cm below the mid-point of the line AB. Calculate the modulus of the string.

In the solution it says resolving vertically gives: 2Tcos& = 0.4g

I don't understand how to resolve vertically or why cos is used, any help??

Reply 1

Hanoi

15

Do you have the Heinemann M3 book? This quesiton is similar to one of the examples in there.

Reply 2

sukhisukhi

OP

ye thats where i got the problem from, trying to follow the example but it doesn't explain how they resolved vertically

Reply 3

ghostwalker

17

sukhisukhi

Two fixed points A and B are on the same horizontal ceiling where AB = 24cm. A light elastic string has one end attached to A and the other end attached to B. The natural length of the string is 20cm. A particle of mass 0.4kg is attached to the mid-point of the string and hangs in equilibrium at a point C, which is 3.5cm below the mid-point of the line AB. Calculate the modulus of the string.

In the solution it says resolving vertically gives: 2Tcos& = 0.4g

I don't understand how to resolve vertically or why cos is used, any help??

Draw the diagram, if you haven't already done so.

To resolve a force in a given direction, you multiply the force by the cosine of the angle between the line of action of the force and the direction you are resolving in.

In this case your mass is in equilibrium, so the sum of the forces acting downwards = the sum of the forces acting upwards.

Acting downwards:

You have just the weight.

Acting upwards:

You have one string going to the right, with force T (tension in the string), and at an angle to the vertical.

Similarly to the left.

And hence your equation.

resolving vertically - mechanics

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