Question

3. A coffee-shop sign is hung from two ropes, each making an angle of 30° with the horizontal (see diagram). The mass of the sign is 12 kg. What is the tension in the left rope?

Answer 1

Hi,

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Please refer to the free body diagram below:

Solve the Tension on the strings (T) into it's perpendicular components as shown in the diagram. T cos 30 degree From figure we have, T sin 30 degree + T sin 30 degree = m g i.e: 2 T sin 30 degree = m = m g We have, Mass of the sign, m = 12 kg Acceleration due to gravity, g = 9.8 m/s^2 therefore Tension on the string is given by, T = m g/2 sin 30 degree = 12 kg times 9.8 m/s^2/2 times 1/2 therefore T = 117.6 N

Resolve the Tension on the strings (T) into it's perpendicular components as shown in the diagram.

$T\cos30^{o}$ will get cancelled each other.

From figure we have,

$T\sin30^{o}+T\sin30^{o}=mg$

$\mathbf{ie;}\;2T\sin30^{o}=mg$

We have,

Mass of the sign, $m=12\;kg$

Acceleration due to gravity, $g=9.8\;m/s^{2}$

$\therefore$ Tension on the string is given by,

$T=\frac{mg}{2\sin 30^{o}}=\frac{12\;kg\times9.8\;m/s^{2}}{2\times \frac{1}{2}}$

$\mathbf{\therefore T=117.6\;N}$

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