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Two friends need to displace a sculpture. They place it on a light board 1.5 m long. One friend lifts at one end with a force of 300 N, and the other friend lifts the opposite end of the board with a force of 160 N. The board is perfectly horizontal. Where along the board is the sculpture center of gravity located? (Tip: The horizontal board has no mass).

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cjmejiab

To solve this problem we will apply the concepts related to the moment. This is defined as the distance of the force to the center of mass (in this case) so the sum of moments must be 0 to maintain the equilibrium condition. The total weight of the sculpture should be

The sculpture must weigh (300 + 160)N = 460 N

From the 300N Force applied, making the summation of moments and fulfilling the condition of static balance, we have to:

[tex]\sum M = 0[/tex]

[tex]160N(1.5m)-460Nx = 0[/tex]

[tex]x = \frac{160(1.5)}{460}[/tex]

x = 0.52m

Therefore the center of gravity is located to 0.52m

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