The firetruck below has a telescoping boom that is used to lift a fireman up in

Question

The firetruck below has a telescoping boom that is used to lift a fireman up in a bucket the boom can be realistically modeled using three length segments and segment #1 is closest to the truck and segment #3 is closest to the bucket the properties of the boom (length, cross-sectional area, and Young's modulus) are given on the sketch. The bucket and fireman weigh 2010 N Find the natural frequency of the system in the vertical direction when the boom is at an angle 50 degree from the horizontal Make a sketch to show how you will model the system Assume that the mass of the telescoping boom is negligible compared to the bucket and fireman and that the boom can only deform in the axial direction (no bending)

Explanation / Answer

GIVEN:- Total mass (M) = 2010N, E = 1.8 x 1011 N/m2, Angle = 50o So total mass accounting along boom = M Sin50, Total length of boom (L) = 7.7m

SOLUTION:- By formula for natural frequency f = 1/2(3.142) x [(3 x E x I) / (MSin50 x L3)] 1/2

Now total inertia of boom I = 0.002 x 3.2Sin(50) + 0.001 x 5.7Sin(50) + 0.0005 x 7.7Sin(50) = 0.012 m4

Substituting above values in frequency formula, we get>

f = 0.16 x [(3 x 1.8 x 1011 x 0.012) / (2010Sin50 x 7.73)]0.5 = 15.36Hz

Hence Natural frequency f = 15.36 Hz (ANSWER)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.