Respuesta :
Answer:
a) weight  of the car = 2816,1 lbs
b) 2773 lbs
Step-by-step explanation:
The equilibrium force is 490 lbs. That force keep the car at rest, then
∑ Fy  =  0   and  ∑Fx  =  0
Forces acting on the car:
The external force  490 lbs
weight of the car  uknown
Normal force
sin∠10°  =  0,174
cos∠10° = 0.985
∑Fx  =  0     mg*sin10°- 490 = 0    ∑Fy =  0    mg*cos10° - N  =  0
mg*0,174= 490
mg  =  490 / 0,174
mg = 2816,1 lbs
weight  of the car = 2816,1 lbs
The Normal force
mg*cos10° - N  =  0     2816,1 * 0,985 = N
N = 2773 lbs
Then equal force in magnitude and in opposite direction will car exets on the driveway
