Respuesta :
Answer:
[tex]x_2=1.60m[/tex]
Explanation:
From the Question We are told that
Initial Force [tex]F_1=5800N[/tex]
Final Force [tex]F_2=6500N[/tex]
Distance between the front and rear wheels \triangle x=3.20 m
Since
[tex]\triangle x=3.20 m[/tex]
Therefore
[tex]x_1+x_2=3.20[/tex]
[tex]x_1=3.20-x_2[/tex]
Generally the equation for The center of mass is at x_2 is mathematically
given by
[tex]x_2 =\frac{(F_1x_1+F_2x_2)}{(F_1+F_2)}[/tex]
[tex]x_2=3.20F_1-\frac{x_2F_1+F_2x_2}{(F_1+F_2)}[/tex]
[tex]2*F_1*x_2 =3.20F_1[/tex]
[tex]x_2=1.60m[/tex]
Center of gravity of a body is the sum of its moments divided by the overall weight of the object. The distance between the front wheels and the truck's center of gravity is 1.6 meters.
Given-
Scale reading value when the front wheels drive over the scale [tex]m_{1}[/tex] is 5800 N.
Scale reading value when the rear wheels drive over the scale [tex]m_{2}[/tex] is 6500 N
Distance between the front and rear wheel [tex]\bigtriangleup x[/tex] is 3.20 meters.
Let, the distance between the front wheels and the truck's center of gravity is [tex]x_{2}[/tex].
Since sum of the distance between front wheel to truck's center of gravity [tex]x_{1}[/tex], and rear wheel to truck's center of gravity [tex]x_{2}[/tex], is equal to the distance between the front and rear wheel [tex]\bigtriangleup x[/tex]. Therefore,
[tex]\bigtriangleup x=x_{1} +x_{2}[/tex]
[tex]3.20=x_{1} +x_{2}[/tex]
[tex]x_{1} =3.20-x_{2}[/tex]
For the distance between the front wheels and the truck's center of gravity is the formula of center of gravity can be written as,
[tex]x_{2} =\dfrac{m_{1}x_{1}+m_{2} x_{2} }{m_{1} +m_{2} }[/tex]
[tex]x_{2} =\dfrac{5800\times (3.20- x_{2})+6500\times x_{2} }{5800 +6500 }[/tex]
[tex]1230 x_{2} ={18560-5800 x_{2}+6500 x_{2} }[/tex]
[tex]x_{2}= 1.6[/tex]
Hence, the distance between the front wheels and the truck's center of gravity is 1.6 meters.
For more about the center of gravity, follow the link below-
https://brainly.com/question/20662119