Question

You pass by a building site and see a crane that lifts loads to upper floors of a building under construction. You wonder about the magnitude of the forces that support the beam of the crane. You snap photograph of the crane with your smartphone. Back at home, you make a drawing of the crank from your photo. Your drawing is shown in the following figure. CM 0 You estimate the lengths and by measure the distance between floors on a point of the building the same distance from your smartphone as the crane and using that distance as a scale for the drawing This tell you that d 100 m and Z - 6.00m. You have drawn the force vector for the gravitational force on the team of the crane, but then you realize you don't know where the center of mass of the beam is located you run back to the construction ice and explain your interest you ask the construction foreman about the crane and he tell you that there itself has a mass of m,- 2.600 kg and the load it waring when you took the photograph has of You then ask the form but the location of the center of mass of the bean Amused by your interest, he contis documents and finds that the center of mass of the cranes located -2.00 horontally to the right of Heals tell you that the pin at Als on a bearing and essentially frictionless, and the point against which the crane pois smooth Tanking the foreman and run home out at your desk and in the forces on the crane at points and a 3) 11 )

Answer 1

We set up the equilibrium conditions as follow:

ΣF, = 0

$F_{Ax}+F_{Bx}=0$

ΣF, = 0

Fay = m19 + m29

$F_{Ay}=(m_{1}+m_{2})g=(2600kg+11850kg)\times 9.8\frac{m}{s^{2}}=141610N$

$\sum M_{A}=0$

$F_{Bx}d=m_{1}gx_{cm}+m_{2}gl$

$F_{Bx}=\frac{(m_{1}x_{cm}+m_{2}l)g}{d}$

$F_{Bx}=\frac{(2600kg\times 2m+11850kg\times 6m)\times 9.8\frac{m}{s^{2}}}{6m}=124623N$

Then from the first equation

$F_{Ax}=-124623N$

In vector notation:

$\bar{F}_{A}=(-124623\hat{i}+141610\hat{j})N$

$\bar{F}_{B}=(124623\hat{i}+0\hat{j})N$