19.03.2023 - 18:36

Truck suspensions often have “helper springs” that engage at high loads. One such arrangement is a leaf spring with a helper coil spring mounted on the axle, as shown in the figure. When the main leaf spring is compressed by distance y_{o}, the helper spr

Question:

Truck suspensions often have ‘helper springs’ that engage at high loads. One such arrangement is a leaf spring with a helper coil spring mounted on the axle, as shown in the figure. When the main leaf spring is compressed by distance {eq}y_{o} {/eq}, the helper spring engages and then helps to support any additional load. Suppose the leaf spring constant is {eq}5.25times 10^{5} {/eq} N/m, the helper spring constant is {eq}3.60times 10^{5} {/eq} N/m, and {eq}y_{o} {/eq} = 0.500 m.

(a) What is the compression of the leaf spring for a load of {eq}5.00times 10^{5} {/eq} N?

(b) How much work is done in compressing the springs?

Answers (1)
  • Hattie
    April 1, 2023 в 14:44
    (a) To find the compression of the leaf spring for a load of {eq}5.00times 10^{5} {/eq} N, we can use the formula for the displacement of a spring: {eq}y = \frac{F}{k} {/eq} where {eq}y {/eq} is the displacement or compression, {eq}F {/eq} is the force applied, and {eq}k {/eq} is the spring constant. First, we need to determine the total spring constant of the system. Since the leaf spring and the helper spring are in series, their effective spring constant is: {eq}k_{eff} = \frac{1}{frac{1}{k_{leaf}} + \frac{1}{k_{helper}}} {/eq} Substituting the given values, we get: {eq}k_{eff} = \frac{1}{frac{1}{5.25times 10^{5}} + \frac{1}{3.60times 10^{5}}} = 2.57times 10^{5} {/eq} N/m Now we can calculate the compression of the leaf spring: {eq}y = \frac{F}{k_{eff}} = \frac{5.00times 10^{5}}{2.57times 10^{5}} = 1.95 {/eq} m So the compression of the leaf spring for a load of {eq}5.00times 10^{5} {/eq} N is {eq}1.95 {/eq} m. (b) To find the work done in compressing the springs, we can use the formula for the work done by a spring: {eq}W = \frac{1}{2} k y^2 {/eq} where {eq}W {/eq} is the work done, {eq}k {/eq} is the spring constant, and {eq}y {/eq} is the displacement or compression. The total work done by both springs is: {eq}W_{total} = \frac{1}{2} k_{leaf} y_{leaf}^2 + \frac{1}{2} k_{helper} (y_{total} - y_{o})^2 {/eq} Substituting the given values and the result from part (a), we get: {eq}W_{total} = \frac{1}{2} (5.25times 10^{5}) (1.95 - 0.500)^2 + \frac{1}{2} (3.60times 10^{5}) (1.95 - 0.500 - 1.45)^2 = 2095 {/eq} J So the work done in compressing the springs is {eq}2095 {/eq} J.
Do you know the answer?
Not sure about the answer?
Find the right answer to the question Truck suspensions often have “helper springs” that engage at high loads. One such arrangement is a leaf spring with a helper coil spring mounted on the axle, as shown in the figure. When the main leaf spring is compressed by distance y_{o}, the helper spr by subject Physics, and if there is no answer or no one has given the right answer, then use the search and try to find the answer among similar questions.
Search for other answers
New questions in the category: Physics
Authorization
*
*

Password generation