A 4 m long, 500 kg steel beam extends horizontally from the point where it is bolted to the frame of a building. A 70 kg man stands on the far end of the beam. What is the magnitude of the torque at the point where the beam attaches to the building

12556.8Nm

Explanation:

A 4 m long, 500 kg steel beam extends horizontally from the point where it is bolted to the frame of a building. A 70 kg man stands on the far end of the beam. What is the magnitude of the torque at the point where the beam attaches to the building

Step one:

Given data

Length of steel x =4m

Mass of steel m1=500kg

Mass of man m2=70kg

The center of the mass of the steel is from the center

Cm=x/2

Cm=4/2= 2m

Torque =force*perpendicular distance

Force =mg

The total torque =torque due to weight of steel +torque due to the man's weight

Total torque = 500*9.81*2+70*9.81*4

Total torque =9810+2746.8

Total torque =12556.8Nm

get the total number of pieces of fruit and divide it by 4 to get the average. the result is the answer to the problem. 14 total number of fruits and there are 4 members. 14 divided by 4 is equals to 3.5.

amplitude

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