Question

1. a car is being pushed on rubber. the force of the push is 4000n. μs for the rubber on the concrete is 1.16. what is the force of gravity on the car? free body diagram given unknown equations substitute/solve 37

Explanation:

Step1: Analyze the situation

When the car is on the verge of moving (maximum static friction case), the pushing force \(F\) is equal to the maximum static - friction force \(F_{s,max}\). The formula for the maximum static - friction force is \(F_{s,max}=\mu_sN\), where \(\mu_s\) is the coefficient of static friction and \(N\) is the normal force. On a horizontal surface, the normal force \(N\) is equal to the force of gravity \(F_g\) (by Newton's second law in the vertical direction \(N - F_g=ma_y\), and \(a_y = 0\) since there is no vertical acceleration).

Step2: Write the equation

We know that \(F = F_{s,max}\), and \(F_{s,max}=\mu_sN\) and \(N = F_g\). So, \(F=\mu_sF_g\).

Step3: Solve for \(F_g\)

We can re - arrange the equation \(F=\mu_sF_g\) to solve for \(F_g\). The formula becomes \(F_g=\frac{F}{\mu_s}\).

Substitute \(F = 4000N\) and \(\mu_s=1.16\) into the formula: \(F_g=\frac{4000N}{1.16}\approx3448.28N\)

Answer:

The force of gravity on the car is approximately \(3448N\)