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9514 1404 393

Step-by-step explanation:

The sum of torques about the pivot point is zero when the system is in equilibrium. That means the total of clockwise torques is equal to the total of counterclockwise torques. For this purpose, torque can be modeled by the product of mass and its distance from the pivot. The uniform beam can be modeled as a point mass at its center.

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a) Let E represent the location of the center of mass of the beam. So, AE = 1.5 m. Then the distance from C to E is AC-AE = AC -1.5 and the CCW torque due to the beam's mass is (16 kg)(AC -1.5 m).

The distance from B to C is 3 m - AC, so the CW torque due to the particle at B is (7 kg)(3 -AC m)

These are equal, so we have ...

  16(AC -1.5) = 7(3 -AC)

  16AC -24 = 21 -7AC . . . . . eliminate parentheses

  23AC = 45 . . . . . . . . . . . add 7AC+24

  AC = 45/23 ≈ 1.957 . . divide by the coefficient of AC

  AC ≈ 2.0 meters . . . . rounded to 1 dp

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b) The torques in this scenario are ...

  M(0.7) = 16(0.8) +7(2.3) . . . . . . AD = 0.7 m, DE = 0.8 m, DB = 2.3 m

  M = 28.9/0.7 ≈ 41.286 . . . . simplify, divide by the coefficient of M

  M = 41.3 kg . . . . rounded to 1 dp

Additional comment

Torque is actually the product of force and distance from the pivot. Here, the forces are all downward, and due to the acceleration of gravity. The gravitational constant multiplies each mass, so there is no harm in dividing the equation by that constant, leaving the sum of products of mass and distance.

a) 0.25x times 3. (could be wrong on equation.)

b) 2.25 pound of dark chocolate. (can i eat all the chocolate? plz? anyway, again, could be wrong, but i hope im right. all the chocolate together should be 5.25 pounds)

have a good day!

answer for #1 is c

answer for #2 is d

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