A pole AB of length 10.0m and weight 600N has its center of gravity 4.0m from the end A, and lies on horizontal ground .Calculate the force required to begin to lift this end

The force required to begin to lift the pole from the end 'A' is 240 N

Explanation:

The given parameters for the pole AB are;

The length of the pole, l = 10.0 m

The weight of the pole, W = 600 N ↓

The distance of the center of gravity of the pole from the side 'A' = 4.0 m

Let '' represent the force required to begin to lift the pole from the end 'A' and let a force applied in the upwards direction be positive

For equilibrium, the sum of moment about the point 'B' = 0, therefore, taking moment about 'B', we have

× 10.0 m - W × 4.0 m = 0

∴ × 10.0 m = W × 4.0 m = 600 N × 4.0 m

× 10.0 m = 600 N × 4.0 m

∴   = 600 N × 4.0 m/(10.0 m) = 240 N

The force required to begin to lift the pole from the end 'A', = 240 N.

i would think form in reponse to tensional forces

hi there!

out of the two, (b) precision would be more reasonable.   however, both a and b could be pretty reasonable.

-awesomerepublic   : )