- #1

MAins

- 18

- 0

Apparently the answer is

∑τ = F_A(1000 m) - mg(20 m) = -2.2x10^9 mN

but I have no idea how to derive this. Please explain!

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- Thread starter MAins
- Start date

- #1

MAins

- 18

- 0

Apparently the answer is

∑τ = F_A(1000 m) - mg(20 m) = -2.2x10^9 mN

but I have no idea how to derive this. Please explain!

- #2

PhanthomJay

Science Advisor

Homework Helper

Gold Member

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- 513

See above in red. If so, calculate the force of the wind (F_A) over the 200m X 70m face of ther building. What is the torque caused by the wind force about the rear edge of the 40 foot wide building? What is the torque of the building's weight about that edge? Which is greater? What does that imply?"A 50 story building is being planned. It is to be 200 m high with a base of 40 m by 70 m. Its total mass will be about 1.8^7 kg, and its weight therefore 1.8 x 10^8 N. Suppose a 200 km/h wind exerts a force of 950 N/m^2 over the 70 m wide face. Calculate the torque about the potential pivot point, the rear edge of the building (where F_E acts in) and determine whether the building will topple. Assume the total force of the wind acts at the midpoint of the building's face, and that the building is not anchored in bedrock."

Apparently the answer is

∑τ = F_A(1000 m) do you mean F_A(100m)? - mg(20 m) = -2.2x10^9 mN

but I have no idea how to derive this. Please explain!

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