Find the tension in the string and the wire?

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Homework Help Overview

The problem involves determining the tension in a string and a wire supporting a microphone. The microphone, with a mass of 500g, is suspended from a wire that is fixed to the ceiling and is held at an angle of 20 degrees to the vertical by a horizontal string. The discussion centers around the equilibrium of forces acting on the microphone.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to convert mass to SI units and the implications of equilibrium on the forces acting on the microphone. There are attempts to equate vertical and horizontal components of tension and weight. Questions arise regarding the calculation of tension in the horizontal string based on the tension in the wire.

Discussion Status

The discussion is ongoing, with participants exploring the relationships between the forces and components of tension. Some guidance has been provided regarding the equilibrium conditions, but there is no explicit consensus on the calculations or final values of the tensions.

Contextual Notes

Participants are working under the assumption that the system is in equilibrium and are discussing the implications of the angle of the wire on the tension calculations. There is also a focus on the need to consider both vertical and horizontal components of the forces involved.

Wayne123
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Find the tension in the string and the wire?

A microphone of mass 500g hangs from the end of a long wire fixed to the ceiling. A horizontal string attached to the microphone exerts a pull which keeps the wire at an angle of 20degrees to the vertical. Find the tensions in both the string and the wire.
 
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Always change to SI units, so 500g = 0.5kg.

Obviously the system is in equilibrium, so all you have to do is to equate the forces.

Vertically:

Force on the microphone due to gravity (mg) = the vertical component of the tension of the long wire.

Horizontally:

The tension in the string = the horizontal component of the tension of the microphone wire.

-----

Remember Horzontal components are bounded by cos θ
Vertical components are bounded by sin θ

where θ = 20
 
Last edited:


i still have a prob. so what ur saying is that:

Tension for the vertical wire= 0.5*10 which is 5N

So what's the tension for the horizontal string?
 


No...

We know the microphone has a weight of 5N, since the system is in equilibrium, this 5N force has to be balanced by the vertical component of Tension in the inclined part of the microphone wire.

Your question states that not all of the microphone is perfectly vertical, imagine half of the wire being straight, whilst the other half if inclined at an angle of 20 degrees to the horizontal - due to the horziontal string.

Now because the wire is inclined, it now has 2 components, a vertical and horizontal component. Obviously the Resultant tension in the wire is T, but we don't know that yet. However we can assume that the vertical component of tension (Tsinθ) is balanced by the weight of the microphone.

So Tsinθ = 5N, rearrange to find the Tension in the wire.

Now we know the tension, we can find the tension in the horizontal string. Because the microphone is not moving side to side, we can assume all horizontal forces are equal.

So the horizontal component of tension in the wire = the tension in the string.

Tcosθ = T(s)

We should know T by now, θ is 20 and we can find T(s)
 

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