Calculating Normal & Friction Forces for 1350kg Car on Plane Surface

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A 1350kg car is resting on a plane surface with locked brakes, and the problem involves calculating the normal and friction forces exerted by the car's wheels. The unit vector normal to the surface is given, along with the tension in a supporting cable. Initial calculations led to an incorrect normal force of 9719N, whereas the expected value is about 2500N higher. The user realized that the components of the frictional vector were misinterpreted, leading to the confusion in calculations. The issue was resolved by correctly relating the components of the frictional vector to the equations.
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1350kg car resting on a plane surface with its brakes locked.
Unit vector e_n = <.231,.923,.308> is perpendicular to the surface. The y-axis points upwards. The direction cosines of a cable supporting the car are <-.816,.408,-.408> and the tension in the cable is 1.2KN. Determine the magnitude of the normal and friction forces the car's wheels exert on the surface.

This is what I have so far:

F_f = <.231F_x, .923F_y, .308F_z>
T_{AB} = 1200<-.816, .408, -.408>
W = -(1350)*(9.8)\hat{j}
N = |1200|<.231,.923,.308>

0 = F_f + T_{AB} + N - W


But when I solve the equations, I come up with N being 9719N, it is supposed to be about 2500 more.

Thanks!
 
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Last edited:
.231N + X -.816*1200=0
.923N+Y-1350*9.8+1200*.408=0
.308N + Z -.408*1200=0
The fourth is correct.
 
Got it

Thanks, I figured this out late last night right before i went to bed.

I forgot that <fx,fy,fz> were already in the form of the frictional vector and that the equation .231Fx + .923Fy + .308Fz was just meant to relate the components, they are not the actual components.

Thanks!
 
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