Question

After driving a portion of the route, the taptap is fully loaded with a total of 27 people including the driver, with an average mass of 69kg per person. In addition, there are three 15-kg goats, five 3-kg chickens, and a total of 25 kg of bananas on their way to the market. Assume that the springs have somehow not yet compressed to their maximum amount. How much are the springs compressed?

Answer 1

We will define the Total mass to calculate the force, so our values are:$M_p = 69*27=1823Kg\\M_g=15*3=45Kg\\M_c=3*5=15Kg\\M_B=25Kg$

Total Mass $= 1863+45+15+25=1948Kg$

The Weight is,

$F=mg=1948*98=19090.4N$

Through the hook's Law we calculate X.

$F_s=Kx$, where x is the lenght of compression and K the Spring constant.

We don't have a K-Spring, but we can assume a random value (or simply let the equation in function of K)

$X = \frac{F_s}{x} \\X = \frac{1909.4}{k}$

I assume a value of $K=4*10^4N/m$

$X= \frac{1909.4}{4*10^4} = 0.48m$