Chapter 9, Problem 69GP

(a)

To determine

To find: The coefficient of static friction ${\mu }_s$ to slide the block.

Answer to Problem 69GP

  ${\mu }_s<\frac{l}{2h}$

Explanation of Solution

Given:

The side of the cube is, $l$

The magnitude of the horizontal force is, $F$

The height from the floor the force is acting is, $h$

The coefficient of static friction is, ${\mu }_s$

Formula used:

Consider the block diagram shown below.

  

From the equilibrium condition in x -direction:

  $\begin{array}{l}{\displaystyle \sum F_x}=0\\ F-F_f=0(1)\\ F_f={\mu }_{s}mg(2)\end{array}$

Consider the torque counter-clockwise as positive. The net torque at point A is zero.

  $\begin{array}{l}{\displaystyle \sum {\tau }_A=0}\\ mg\left(\frac{l}{2}\right)-F\times h=0\left(3\right)\end{array}$

Calculation:

From equation (1) and (2),

  $F={\mu }_{s}mg(4)$

Substitute the value of $F$ in equation (3).

  $\begin{array}{l}mg\left(\frac{l}{2}\right)-{\mu }_{s}mg\times h=0\\ {\mu }_s=\frac{l}{2h}\end{array}$

For the block began to slide rather than a tip.

  ${\mu }_s<\frac{l}{2h}$

Conclusion:

The value of static friction should be less than $\frac{l}{2h}$ .

(b)

To determine

To find: The coefficient of static friction ${\mu }_s$ to block to tip.

Answer to Problem 69GP

  ${\mu }_s>\frac{l}{2h}$

Explanation of Solution

Given:

The side of the cube is, $l$

The magnitude of the horizontal force is, $F$

The height from the floor the force is acting is, $h$

The coefficient of static friction is, ${\mu }_s$

Formula used:

Consider the block diagram shown below.

  

From the equilibrium condition in x -direction:

  $\begin{array}{l}{\displaystyle \sum F_x}=0\\ F-F_f=0(1)\\ F_f={\mu }_{s}mg(2)\end{array}$

Consider the torque counter-clockwise as positive. The net torque at point A is zero.

  $\begin{array}{l}{\displaystyle \sum {\tau }_A=0}\\ mg\left(\frac{l}{2}\right)-F\times h=0(3)\end{array}$

Calculation:

From equation (1) and (2).

  $F={\mu }_{s}mg(4)$

Substitute the value of $F$ in equation (3).

  $\begin{array}{l}mg\left(\frac{l}{2}\right)-{\mu }_{s}mg\times h=0\\ {\mu }_s=\frac{l}{2h}\end{array}$

For the block began to tip,

  ${\mu }_s>\frac{l}{2h}$

Conclusion:

The value of static friction should be more than $\frac{l}{2h}$ .