What Tension is Required for a 10 kg Mass to Hang Motionless?

[WMDhamnekar]

Gravity acting on a 10 kg,mass produces a force of $F_g=\langle 0, -98\rangle$ Newtons. If the mass is suspended from 2 wires which both form $30^\circ$ angles with the horizontal, then what forces of tension are required in order for the mass to hang motionless over time?

Answer.
I computed $F_{net}=\langle 0,-98+F\rangle$. Now how to compute the final answer? For equilibrium,$F_{net}=0$

 

[skeeter]

Gravity acting on a 10 kg,mass produces a force of $F_g=\langle 0, -98\rangle$ Newtons. If the mass is suspended from 2 wires which both form $30^\circ$ angles with the horizontal, then what forces of tension are required in order for the mass to hang motionless over time?

Answer.
I comuted $F_{net}=\langle \frac {\sqrt{3}-1}{2},-98+\frac{(\sqrt{3}+1)*F}{2}\rangle$. Now how to compute the final answer? For equilibrium,$F_{net}=0$

sum of the magnitudes of each vertical component of tension upward = magnitude of gravity force downward

$2T\sin(30) = mg$

$T = \dfrac{mg}{2\sin(30)} = mg$

 

[WMDhamnekar]

sum of the magnitudes of each vertical component of tension upward = magnitude of gravity force downward

$2T\sin(30) = mg$

$T = \dfrac{mg}{2\sin(30)} = mg$

Hello,

So $F=\frac{98}{\sqrt{2}}=69.2965N$ approx. Isn't it?

 

[skeeter]

Hello,

So $F=\frac{98}{\sqrt{2}}=-69.2965$ approx. Isn't it?

No ... magnitude of tension in each wire oriented 30 degrees from the horizontal is $T = mg = 98 \text{ N}$

As the angle from the horizontal increases, the tension decreases. If both wires are vertical ($\theta = 90^\circ$), the tension is a minimum ...

$T = \dfrac{mg}{2\sin(90)} = \dfrac{98}{2} = 49 \text{ N}$

 

[WMDhamnekar]

No ... magnitude of tension in each wire oriented 30 degrees from the horizontal is $T = mg = 98 \text{ N}$

As the angle from the horizontal increases, the tension decreases. If both wires are vertical ($\theta = 90^\circ$), the tension is a minimum ...

$T = \dfrac{mg}{2\sin(90)} = \dfrac{98}{2} = 49 \text{ N}$

Is my computed answer for $F_{net}$ is correct?

 

[skeeter]

Is my computed answer for $F_{net}$ is correct?

The system is in equilibrium ...

$\displaystyle \sum F_x = 0 \text{ and } \sum F_y = 0 \implies F_{net} = 0$

 

[WMDhamnekar]

The system is in equilibrium ...

$\displaystyle \sum F_x = 0 \text{ and } \sum F_y = 0 \implies F_{net} = 0$

So, the final answer for the forces of tension required in order for the mass to hang motionless over time is 98N.

 

[skeeter]

So, the final answer for the forces of tension required in order for the mass to hang motionless over time is 98N.

does the attached sketch confirm your statement?

View attachment 8835