Question

1. in the rover below, calculate the x - y coordinates of the center of gravity in reference to the origin.

the wheelbase is 12 in and the track width is 8 in. the x - direction is left to right, and the y - direction is
up and down. the front of the rover is to the right and the origin is located at the center of the
right rear wheel. draw the free body diagrams.
1.15lbf 12 in 0.75lbf
8in.
origin
1.25lbf 0.65lbf

Explanation:

Step1: Calculate the total force

The total force \(F_{total}\) is the sum of all the forces:
\(F_{total}=1.15 + 0.75+1.25 + 0.65=\) \(3.8\) lbf

Step2: Calculate the \(x -\) coordinate of the center of gravity

The formula for the \(x -\) coordinate of the center of gravity \((x_{cg})\) is \(\sum(F_i\times x_i)/F_{total}\).
For the left - side forces (left - rear: \(F_1 = 1.25\) lbf, \(x_1 = 0\); left - front: \(F_2=1.15\) lbf, \(x_2 = 12\)) and right - side forces (right - rear: \(F_3 = 0.65\) lbf, \(x_3 = 0\); right - front: \(F_4 = 0.75\) lbf, \(x_4 = 12\))
\(x_{cg}=\frac{(1.25\times0 + 1.15\times12+0.65\times0 + 0.75\times12)}{3.8}\)
First, calculate the numerator: \(1.15\times12+0.75\times12=(1.15 + 0.75)\times12=1.9\times12 = 22.8\)
Then \(x_{cg}=\frac{22.8}{3.8}=6\) in

Step3: Calculate the \(y -\) coordinate of the center of gravity

The formula for the \(y -\) coordinate of the center of gravity \((y_{cg})\) is \(\sum(F_i\times y_i)/F_{total}\).
For the rear - side forces (left - rear: \(F_1 = 1.25\) lbf, \(y_1 = 0\); right - rear: \(F_3 = 0.65\) lbf, \(y_3 = 0\)) and front - side forces (left - front: \(F_2=1.15\) lbf, \(y_2 = 8\); right - front: \(F_4 = 0.75\) lbf, \(y_4 = 8\))
\(y_{cg}=\frac{(1.25\times0+0.65\times0 + 1.15\times8+0.75\times8)}{3.8}\)
First, calculate the numerator: \(1.15\times8+0.75\times8=(1.15 + 0.75)\times8=1.9\times8 = 15.2\)
Then \(y_{cg}=\frac{15.2}{3.8}=4\) in

Answer:

The \(x -\) coordinate of the center of gravity is \(6\) in and the \(y -\) coordinate of the center of gravity is \(4\) in.