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- Homework Statement:
- Shown in the figure below are two light plastic supports pinned at their top. One of the supports is 4 m long while the other is 3 m long and they are arranged to form a right angled triangle with their base 5 m. A mass of 300 N is hung from the pin at the top. If the ground is frictionless, calculate the reactions ##n_1## and ##n_2## at the bases of the two supports.

- Relevant Equations:
- For equilibrium, ##\Sigma \vec F = 0## for the system as a whole and ##\Sigma \vec \tau = 0## for any point on the system where torque ##\vec \tau = \vec r \times \vec F##.

For equilibrium, using ##\Sigma \vec F = 0##, we get ##n_1 + n_2 = 300\; \text{N}##.

Taking the system as a whole and applying ##\Sigma \vec \tau = 0## about the hinge (pin) at the top from where the load is hung, we get ##n_1 \times (0.8) \times 4 = n_2 \times (0.6) \times 3##, by taking those components of the normal forces perpendicular to the supports and using trigonometry.

Hence, ##3.2 n_1 = 1.8 n_2 \Rightarrow n_2 = \frac{16}{9} n_1##.

Thus, going to the earlier equation, ##n_1 + \frac{16}{9} n_1 = 300 \Rightarrow \frac{25}{9} n_1 = 300 \Rightarrow \boxed{n_1 = 108 \; \text{N}}##.

Also, ##\boxed{n_2 = 192 \; \text{N}}##.

**Is this right?**

Even if it is, is there a physical explanation as to why ##n_2 > n_1##? Can we say it is because ##n_2## has a smaller lever arm than ##n_1## and therefore has to be greater in order to nullify the torque produced by ##n_1## about the hinge?