Question

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the fbd in figure 8 is for a car at rest on the ground. t/i
(a) determine fs 1 and fs g. ans: f 1 = 2650 n r; f g = 13 000 n down
(b) how would your answers change if the car was moving at a constant velocity? assume none of the given forces change

Explanation:

Step1: Analyze forces in horizontal direction for part (a)

For the horizontal forces, using the equilibrium condition \(F_{net - x}=0\). The sum of the forces to the left is \(1250\ N + 1400\ N=2650\ N\). So, \(F_1 = 1250\ N+1400\ N\).

Step2: Analyze forces in vertical direction for part (a)

For the vertical forces, using the equilibrium condition \(F_{net - y}=0\). Since the normal force \(N = 13000\ N\) (upward) and the car is at rest, \(F_g=N\) (by Newton's first law, when \(a = 0\), \(F_{net}=0\)). So \(F_g=13000\ N\) (downward).

Step3: Analyze the case when the car is moving at constant velocity (part (b))

When the car is moving at a constant velocity (\(a = 0\)), according to Newton's first law \(F_{net}=ma\), and since \(a = 0\), \(F_{net}=0\). The equilibrium conditions \(F_{net - x}=0\) and \(F_{net - y}=0\) still hold. The forces in the horizontal and vertical directions are balanced in the same way as when the car is at rest.

Answer:

(a) \(F_1 = 2650\ N\) (to the right) and \(F_g = 13000\ N\) (downward).
(b) The answers would not change. When the car is moving at a constant velocity, the net force \(F_{net}=0\) (Newton's first law: \(F_{net}=ma\) and \(a = 0\)). So the balance of forces (equilibrium conditions \(F_{net - x}=0\) and \(F_{net - y}=0\)) is the same as when the car is at rest.