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Chapter 10: Q96P (page 293)

Beverage engineering. The pull tab was a major advance in the engineering design of beverage containers. The tab pivots on a central bolt in the can’s top. When you pull upward on one end of the tab, the other end presses downward on a portion of the can’s top that has been scored. If you pull upward with a10 N force, approximately what is the magnitude of the force applied to the scored section? (You will need to examine a can with a pull tab.)

Short Answer

Expert verified

The magnitude of forcethat acts on the scored sectionis25 N .

Step by step solution

01

Step 1: Given

i) Distances,r1=1.8cm,  r2=0.73cm

ii) Force,F1=10 N

02

Determining the concept

As the tab is in equilibrium, the net torque exerted by both forces must be equal. Therefore, by using the formula of torque, find the magnitude of force that acts on the scored section.

The formula is as follows:

τup=τdown

Where, τ is torque

03

Determining themagnitude of force that acts on the scored section

For torque in equilibrium,

τup=τdown

F1r1=F2r2

10 N×1.8 m=F2×0.73 m

F2=(1.8 cm0.73 cm)×(10 N)

F2=25 N

Hence, themagnitude of force that acts on the scored section is25 N .

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Most popular questions from this chapter

Two uniform solid spheres have the same mass of1.65 kg, but one has a radius of0.226 mand the other has a radius of. Each can rotate about an axis through its center. (a) What is the magnitudeof the torque required to bring the smaller sphere from rest to an angular speed of317 rad/sin15.5 s? (b) What is the magnitudeof the force that must be applied tangentially at the sphere’s equator to give that torque? What are the corresponding values of (c)τand (d)for the larger sphere?

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