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?

Enter the compression numerically in meters using two significant figures.

compression of spring is x = 0.12 m

Assumed k = 160,000 N/m ... Truck's suspension system

Explanation:

Given:

- The mass of average person m_p = 69 kg

- Total number of persons n_p = 27

- The mass of each goat m_g = 15 kg

- The total number of goats n_g = 3

- The mass of each chicken m_c = 3 kg

- The total number of goats n_c = 5

- The total mass of bananas m_b = 25 kg

Find:

How much are the springs compressed?

Solution:

- Using equilibrium equation on the taptap in vertical direction:

F_net = F_spring - F_weight = 0

- Compute the force due to all the weights on the taptap:

F_weight = (n_p*m_p + n_g*m_g + n_c*m_c + m_b) * 9.81

F_weight = (69*27 + 3*15 + 5*3 + 25) * 9.81

F_weight = 19109.88 N

- The restoring force of a spring is given by:

F_spring = k*x

Where, k is the spring stiffness and x is the displacement:

F_weight = F_spring

19109.88 = k*x

x = 19109.88 / k

We need to assume the spring stiffness we will take k = 160,0000 N/m (trucks suspension systems). The value of the stiffness must be high enough to sustain a load of 1.911 tonnes.

x = 19109.88 / 160,000

x = 0.1194 m ≈ 0.12 m = 12 cm

- A compression of 12 cm seems reasonable for a taptap to carry 1.911 tonnes of load. Hence, the assumption of spring stiffness was reasonable. Hence, the compression of spring is x = 0.12 m.