You hang a 3 kg Halloween decoration 21 cm from the right end of a curtain rod. How much force does one of the curtain rod supports need to be able to hold? The supports are at the very ends of the rod, which has mass 2.1 kg and is 114 cm long. Answer in newtons.

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JemdetNasr JemdetNasr · Answer 1 · 2019-07-20T10:53:46+02:00

Answer:

34.3 N and 15.7 N

Explanation:

[tex]F_{left}[/tex] = force on the left end of the rod

[tex]F_{right} [/tex] = force on the right end of the rod

M = mass of Halloween decoration = 3 kg

[tex]F_{h}[/tex] = weight of the Halloween decoration = Mg = 3 x 9.8 = 29.4 N

m = mass of rod = 2.1 kg

[tex]F_{r}[/tex] = weight of the rod = mg = 2.1 x 9.8 = 20.6 N

From the force diagram, using equilibrium of torque about A

[tex]F_{h}[/tex] (AB) + [tex]F_{r}[/tex] (AC) = [tex]F_{right}[/tex] (AD)

(29.4) (21) + (20.6) (57) = [tex]F_{right}[/tex] (114)

[tex]F_{right}[/tex] = 15.7 N

Using equilibrium of force along the vertical direction

[tex]F_{right}[/tex] + [tex]F_{leftt}[/tex] = [tex]F_{h}[/tex] + [tex]F_{r}[/tex]

15.7 + [tex]F_{leftt}[/tex] = 29.4 + 20.6

[tex]F_{leftt}[/tex] = 34.3 N