Question

A horizontal rod with a mass of 10 kg and length 12 m is hinged to a wall at one end
and supported by a cable which makes an angle of 30º with the rod at its other end.
Calculate the tension in the cable and the force exerted by the hinge.

Answer 1

Answer on Question 73158, Physics, Mechanics, Relativity

Question:

A horizontal rod with a mass of $10\,kg$ and length $12\,m$ is hinged to a wall at one end and supported by a cable which makes an angle of $30{}^\circ$ with the rod at its other end. Calculate the tension in the cable and the force exerted by the hinge.

Solution:

a) The sum of torques about the hinge must be equal to zero:

$$\begin{array}{l} \sum \tau = 0, \\ - W \frac {L}{2} + T L \sin \alpha = 0, \\ - m g \frac {L}{2} + T L \sin \alpha = 0, \\ m g \frac {L}{2} = T L \sin \alpha , \\ \end{array}$$

here, $W = mg$ is the weight of the rod, $m = 10 \, kg$ is the mass of the rod, $g = 9.8 \, m/s^2$ is the acceleration due to gravity, $L = 12 \, m$ is the length of the rod, $T$ is the tension in the cable and $\alpha = 30{}^\circ$ is the angle which the cable makes with the rod.

Then, from this formula we can find the tension in the cable:

$$T = \frac {m g}{2 \sin \alpha} = \frac {1 0 k g \cdot 9 . 8 \frac {m}{s ^ {2}}}{2 \cdot \sin 3 0 {}^ {\circ}} = 9 8 N.$$

b) The sum of the forces must be equal to zero:

$$\sum F = 0.$$

Let's write the sum of the components of the forces in projections on axis $x$ and $y$ :

$$F _ {x} - T _ {x} = 0,$$ $$F _ {x} = T _ {x} = T \cos \alpha = 9 8 N \cdot \cos 3 0 {}^ {\circ} = 8 5 N,$$ $$F _ {y} + T _ {y} - W = 0,$$ $$F _ {y} = m g - T \sin \alpha = 1 0 k g \cdot 9. 8 \frac {m}{s ^ {2}} - 9 8 N \cdot \sin 3 0 {}^ {\circ} = 4 9 N.$$

Finally, we can find the force exerted on the hinge from the Pythagorean theorem:

$$F = \sqrt {F _ {x} ^ {2} + F _ {y} ^ {2}} = \sqrt {(8 5 N) ^ {2} + (4 9 N) ^ {2}} = 9 8 N.$$

Answer:

a) $T = 98N$

b) $F = 98N$

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