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Chapter 9: Q2PE (page 316)

When tightening a bolt, you push perpendicularly on a wrench with aforce of 165 N at a distance of 0.140 m from the center of the bolt. (a)How much torque are you exerting in newton × meters (relative to thecenter of the bolt)? (b) Convert this torque to foot-pounds.

Short Answer

Expert verified

(a)The torque is\(23.1\;{\rm{N}} \cdot {\rm{m}}\).

(b)The torque is\(17.4\;{\rm{foot}} \cdot {\rm{poundal}}\).

Step by step solution

01

Step 1:Given data and torque calculation

  • The force is\(F = 165\;{\rm{N}}\).
  • The perpendicular distance from the hinge is\(r = 0.140\;{\rm{m}}\).

The torque is,

\(\tau = r \times F\)

Here\(r\)is the perpendicular distance and substituting the values we get,

The torque

\(\begin{array}{c}\tau = 0.140 \times 165\\ = 23.1\;{\rm{N}} \cdot {\rm{m}}\end{array}\)

02

Step 2:Calculation of torque in the foot-poundal

(b)

The torque is\(23.1\;{\rm{N}} \cdot {\rm{m}}\).

We know,

\(1\;{\rm{m}} = 3.28\;{\rm{feet}}\)

\(1\;{\rm{N}} = 0.225\;{\rm{pound}}\)

So,

\(\begin{array}{c}\tau = 23.1\;{\rm{N}} \cdot {\rm{m}}\\ = 23.1 \times 0.225 \times 3.28\\ = 17.04\;{\rm{foot}} \cdot {\rm{poundal}}\end{array}\)

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