Chapter 8, Problem 26P

(a)

To determine

To find: The force that is perpendicular to the wrench based on the mechanic exert.

Answer to Problem 26P

The force value is $314\text{N}$ .

Explanation of Solution

Given information:

The bolts on the cylinder head of an engine require tightening to a torque of $88\text{N}\cdot \text{m}$ . If a wrench is $28\text{cm}$ long, and the six-sided bolt head is $15\text{mm}$ in diameter.

Formula used:

$\tau =rF\sin \theta$

Here,

$\tau$ is the torque,

$r$ is the distance ,

$\theta$ is the angle,

$F$ is the force.

Calculation:

The torque applied $\left(\tau \right)=88\text{N}\cdot \text{m}$

Length of wrench $\left(r\right)=0.28\text{m}$

The force needed to produce the required toque

$\tau =rF\sin \theta$

Force is applied perpendicular to the wrench

So, angle $\theta =90°$

$\begin{array}{l}F=\tau /r\\ =88/0.28\\ F=314\text{}\mathrm{N}\end{array}$

Conclusion:

The force value is 314N.

(b)

To determine

To find: The force that is exerted based on each six points.

Answer to Problem 26P

The value is $F_{\text{bolt}}=2.0\times {10}^3\text{N}$ .

Explanation of Solution

Given information:

The bolts on the cylinder head of an engine require tightening to a torque of $88\text{N}\cdot \text{m}$ . If a wrench is $28\text{cm}$ long, and the six-sided bolt head is $15\text{mm}$

in diameter.

Formula used:

$F_{\text{bolt}}=\left(\frac{\left(\tau /r\right)}{6}\right)$

Here,

$\tau$ is the torque,

$r$ is the distance .

Calculation:

Diameter of bolt head $=15\times {10}^{-3}\text{m}$

The net torque on the bolt is same

Because this torque is balanced by the torque produced by the bolt, an equal torque is produced on the bolt

This torque is produced by six forces, one at the each of six points

$\begin{array}{l}{F}_{\text{bolt}}=\left(\frac{\left(\tau /r\right)}{6}\right)\\ =\left(\frac{88}{6\times 7.5\times {10}^{-3}}\right)\text{N}\\ F_{\text{bolt}}=2.0\times {10}^3\text{N}\end{array}$

Conclusion:

The value is $F_{\text{bolt}}=2.0\times {10}^3\text{N}$ .