Chapter 9, Problem 1P

To determine

Magnitude of static frictional force.

Answer to Problem 1P

842.857 N

Explanation of Solution

Given:

Radius of the wheel of the car $r=0.350\text{ m}$

Torque applied by the engine of the car $\tau =295\text{ N}\cdot \text{m}$

Formula used:

  $\begin{array}{l}\tau =\overrightarrow{r}\times \overrightarrow{F}\\ \tau =rF\sin \theta \\ \tau =torque\\ r=radius\\ F=force\\ \theta =angle\end{array}$

Calculation:

From the given question when the car engine applies the torque to the wheel of the car, it does not slip against the road surface.

Hence, the frictional torque should be equal to the applied torque.

  $\begin{array}{l}\tau =\overrightarrow{r}\times \overrightarrow{F}\\ \tau =rF\sin \theta \\ \tau =rF\sin {90}^{\circ }\\ \tau =0.350*F*1\\ \tau =0.350F\\ F=\frac{\tau }{0.350}\\ F=\frac{295}{0.350}\\ F=842.857\text{ N}\end{array}$

Conclusion:

Therefore, the static frictional force is $842.857\text{ N}$