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Answer :
To find the force a curtain rod support must hold, calculate the torques due to the rod's weight and the decoration's weight, considering their distances from the support. Then, balance these torques with the force at the other support. The calculated force for one support is approximately 15.7 newtons.
To calculate the force one of the curtain rod supports must be able to hold, we must consider the torques about one end of the rod and include the mass of the rod and the Halloween decoration. The rod and the decoration together exert a torque about the pivot point due to gravity. The support must exert an equal and opposite torque to keep the system in equilibrium. We do this by summing the torques and setting them to zero, since the system is not rotating.
The total torque ( au) about the right support due to the rod and the decoration can be calculated using au = r imes F where r is the distance from the pivot and F is the force due to gravity (weight). The weight of the rod acts at its center of mass, which is at the midpoint of the rod, while the weight of the decoration acts at 21 cm from the right end. So, the distances from the pivot (right end) to the centers of mass are:
Rod: 57 cm (half of the total length of 114 cm)
Decoration: 21 cm
The weights are:
Rod: 2.1 kg imes 9.8 m/s2 = 20.58 N
Decoration: 3 kg imes 9.8 m/s2 = 29.4 N
The torques about the right support are:
Rod: 57 cm imes 20.58 N = 1173.06 N ext{cm}
Decoration: 21 cm imes 29.4 N = 617.4 N ext{cm}
Since the rod is in equilibrium, the torques must balance, meaning the support at the left end must provide an upwards force resulting in a torque that equals the sum of the other two torques. Let F be the force at the left end support. It must satisfy:
F imes 114 cm = 1173.06 N ext{cm} + 617.4 N ext{cm}
F = (1173.06 + 617.4) / 114
F ≈ 15.7 N
Therefore, the force one of the curtain rod supports needs to be able to hold is approximately 15.7 newtons.
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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
