Answer:
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.
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