Chapter 5, Problem 10STP

To determine

The magnitude of the static friction force acting on the boy.

Answer to Problem 10STP

The magnitude of the static friction force acting on the boy is $f_{\text{static}}=178\text{N}$ .

Explanation of Solution

Given:

The weight of the person S is $41.2\text{kg}$ .

The person J applies extremely sticky paste to the top of the banister.

The coefficient of static friction of the banister is $0.72$ .

The angle between the banister and the horizontal is $52.4°$ .

Formula used:

Consider a force $F$ is acting on an object at an angle $\theta$ with the horizontal. The magnitude of horizontal component of the force $F$ is,

  $F_x=F\cos \theta$

The magnitude of vertical component of the force $F$ is,

  $F_y=F\sin \theta$

The expression for the static friction $\left(f_{\text{static}}\right)$ acting on the object is,

  $f_{\text{static}}=\mu F_{\text{N}}$

Here, $\mu$ is the coefficient of static friction and $F_{\text{N}}$ is the normal force acting on the object.

Calculation:

The free body diagram of the situation is shown below.

To find the normal force $\left(F_{\text{N}}\right)$ acting on the person S , equate the forces acting in the $y$ direction,

$\begin{array}{l}{F}_{\text{N}}=mg\cos \theta \\ F_{\text{N}}=\left(41.2\text{kg}\right)\times \left(9.80{\text{m/s}}^2\right)\times \cos \left(52.4°\right)\\ F_{\text{N}}=\left(41.2\text{kg}\right)\times \left(9.80{\text{m/s}}^2\right)\times \left(0.61014\right)\\ F_{\text{N}}=246.3\text{kg}\cdot {\text{m/s}}^{\text{2}}\times \left(\frac{\text{1}\text{N}}{1\text{kg}\cdot {\text{m/s}}^{\text{2}}}\right)\\ F_{\text{N}}=246.3\text{N}\end{array}$

The static frictional force acting on the person S is,

$\begin{array}{l}{f}_{\text{static}}=\mu F_{\text{N}}\\ f_{\text{static}}=0.72\times 246.3\\ f_{\text{static}}=178\text{N}\end{array}$

Conclusion:

Thus, the magnitude of the static friction force acting on the boy is $f_{\text{static}}=178\text{N}$ .