# Video: Determining the Compression of Loaded Springs

The springs of a pickup truck act like a single spring with a force constant of 1.30 × 10⁵ N/m. By how much will the truck be depressed by its maximum load of 1.00 × 10³ kg? If the pickup truck has four identical springs, what is the force constant of each spring?

02:32

### Video Transcript

The springs of a pickup truck act like a single spring with a force constant of 1.30 times 10 to the fifth newton metres. By how much will the truck be depressed by its maximum load of 1.00 times 10 to the third kilograms? If the pickup truck has four identical springs, what is the force constant of each spring?

In the first part of this exercise, when we’re asked by how much the truck is depressed, we’re searching for a displacement which we can call Δ𝑥. In part two where we want to solve for the force constant of each of the four identical springs, we can call that force constant 𝑘 sub four. To start off on our solution, we can recall Hooke’s law for springs. This law tells us that a spring’s restoring force is equal to negative its spring constant multiplied by its displacement from equilibrium Δ𝑥. In the case of our pickup truck that’s fully loaded, we can call the mass of that load 𝑚 given as 1.00 times 10 to the third kilograms. And we’re told that the overall spring constant of the truck’s suspension system, which we’ve called 𝑘, is 1.30 times 10 to the fifth newton metres.

Since the fully loaded truck is in equilibrium, we can write that the magnitude of the gravitational force on the full load of the truck is equal to the magnitude of the restoring spring force or 𝑚 times 𝑔 equals 𝑘 times Δ𝑥. And when we rearrange for Δ𝑥, we find it’s equal to 𝑚 times 𝑔 over the spring constant 𝑘. The acceleration due to gravity 𝑔 we’ll let be exactly 9.8 metres per second squared. When we plug in for 𝑚, 𝑔, and 𝑘 and enter this expression on our calculator, we find Δ𝑥 is 7.54 centimetres. That’s how much the springs of the truck’s suspension system are compressed under this load.

Next, we want to solve for the spring constant if rather than one collective spring constant for the suspension system, it was equally divided among the four wheels of the truck. Since there is an equal division, that means that 𝑘 sub four is equal to 𝑘 the spring constant of the truck collectively divided by four. Plugging in for 𝑘 to three significant figures, we find a value of 3.25 times 10 to the fourth newton metres. That’s the effective spring constant of each of the truck’s wheels suspension systems.