Answer to Question #157710 in Physics for Andu

Question

Calculate the weight of a machine with mass m = 1.2 t when it passes with velocity v = 20 m / s through the lowest point of a bridge bent down. The radius of the bridge is R = 100 m. Compare this weight with the weight of the car at rest. What is the cause of the change in weight of the car compared to the usual weight?

Answer 1

The weight of the car at rest can be found as follows:

$$W_{rest}=F_N=mg=1.2\cdot10^3\ kg\cdot9.8\ \dfrac{m}{s^2}=11760\ N.$$

The weight of the car at the lowest point of a bridge bent down can be found as follows:

$$W=F_N=\dfrac{mv^2}{R}+mg,$$ $$W=F_N=\dfrac{1.2\cdot10^3\ kg\cdot(20\ \dfrac{m}{s})^2}{100\ m}+11760\ N=16560\ N.$$

Let's compare this weight with the weight of the car at rest:

$$\dfrac{W}{W_{rest}}=\dfrac{16560\ N}{11760\ N}=1.4.$$

As we can see from calculations, the weight of the car at the lowest point of a bridge bent down 1.4 times larger than the weight of the car at rest. The centripetal acceleration cause the change in weight of the car compared to the usual weight.