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wkfrst
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I know I am missing something simple here. This is only the second problem and I have done the rest with no problems. I would appreciate a second set of eyes.
The mass center G of the 1430-kg rear-engine car is located as shown in the figure. Determine the normal force under the left front tire (a)Nf (point F) and under the left rear tire (b)Nr (point R) when the car is in equilibrium.
F is the front tire. Center of gravity is 1406mm to the right of F. R is rear tire which is 972mm from CG or 2378 to the right of F.
I know you can use 3 different equations to solve these 2D static problems. Sum of forces in x direction, sum of forces in y direction, and sum of the moment about a point of your choosing.
All the forces are normal to the x-axis so there are no forces acting in the x direction. Summing in the y-direction gives you two unknowns of F(front) and R(rear).
I chose to sum the moment about F. The normal force at F slides through the point so the it's contribution is zero.
M about F= (9.81m/s^2)(1430kg)(1.406m)-R(2.378m)=0
I chose clockwise to be positive.
The answer I get for R is 8294N but apparently is wrong. I am guessing I am over looking something simple.
Thanks for the help!
Homework Statement
The mass center G of the 1430-kg rear-engine car is located as shown in the figure. Determine the normal force under the left front tire (a)Nf (point F) and under the left rear tire (b)Nr (point R) when the car is in equilibrium.
F is the front tire. Center of gravity is 1406mm to the right of F. R is rear tire which is 972mm from CG or 2378 to the right of F.
Homework Equations
I know you can use 3 different equations to solve these 2D static problems. Sum of forces in x direction, sum of forces in y direction, and sum of the moment about a point of your choosing.
The Attempt at a Solution
All the forces are normal to the x-axis so there are no forces acting in the x direction. Summing in the y-direction gives you two unknowns of F(front) and R(rear).
I chose to sum the moment about F. The normal force at F slides through the point so the it's contribution is zero.
M about F= (9.81m/s^2)(1430kg)(1.406m)-R(2.378m)=0
I chose clockwise to be positive.
The answer I get for R is 8294N but apparently is wrong. I am guessing I am over looking something simple.
Thanks for the help!
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