Chapter 10.2, Problem 25PP

To determine

To identify: the quantity which changes if the radius is doubled.

how much changes are made if the radius is doubled.

Explanation of Solution

Given:

Radius is doubled.

Radius of the cycle is 35.6 cm.

Effort force applied on the chain is 155 N.

Efficiency of the bicycle is 95 %.

Distance moved by applying the effort force is 14 cm.

Formula used:

The ideal mechanical advantage for an ideal machine in terms of radius is equal to the ratio of the radius of the effort force to the radius of the load and is given by

  $IMA=\frac{r_e}{r_l}$  ......(1)

Here, $r_e$ and $r_l$ are the displacement of the effort force and $r_l$ is the displacement of the load.

The efficiency of a machine ( in percentage %) is the ratio of its mechanical advantage to the ideal mechanical advantage and multiply by 100 ,that is,

  $e=\left(\frac{MA}{IMA}\right)100$  ......(2)

Here, $MA$ is the mechanical advantage of a machine and $IMA$ is the ideal mechanical.

Calculation:

Mechanical advantage of a machine is equal to the ratio of the resistance force to the effort force and is given by

  $MA=\frac{F_r}{F_e}$  ......(3)

Here, $F_e$ and $F_r$ are the effort force and resistance force.

The ideal mechanical advantage for an ideal machine in the terms of distance is equal to the ratio of the displacement of the effort force to displacement of the load and is given by

  $IMA=\frac{d_e}{d_l}$  ......(4)

Here, $d_e$ and $d_r$ are the displacement of the effort force and displacement of the load.

The double value of the gear radius of the cycle is $2\left(4\text{cm}\right)=8\text{cm}$ .

Solve for $IMA$

Substitute the values of $r_e$ and $r_r$ in equation (1)

  $\begin{array}{l}IMA=\frac{8\text{cm}}{35.6\text{cm}}\\ \boxed{IMA=0.225}\end{array}$

Therefore, the value of ideal mechanical advantage is 0.225.

Solve for $MA$

Rearrange the equation (2) for $MA$

  $MA=\left(\frac{e}{100}\right)\times \left(IMA\right)$  ......(5)

Substitute the values of $e$ and $IMA$ in equation (5),

  $\begin{array}{l}MA=\frac{95}{100}\times 0.225\\ \boxed{MA=0.214}\end{array}$

Therefore, the value of mechanical advantage is 0.214.

Solve for the resistance force $F_r$

Rearrange the equation (3) for $F_r$

  $F_r=\left(MA\right)F_e$  ......(6)

Substitute the values of $MA$ and $F_e$ in equation (6)

  $\begin{array}{c}{F}_r=\left(MA\right)F_e\\ =\left(0.214\right)\left(155\text{N}\right)\\ \boxed{F_r=33.2\text{N}}\end{array}$

Therefore, the round off value up to two digits of resistance force is $32\text{N}$ .

Solve for the distance $d_e$

Rearrange the equation (4)

  $d_e=\left(IMA\right)\left(d_r\right)$  ......(7)

Substitute the value of $IMA$ and $d_r$ in equation (7)

  $\begin{array}{c}{d}_e=\left(IMA\right)\left(d_r\right)\\ =\left(0.224\right)\left(14\text{cm}\right)\\ \boxed{d_e=3.13\text{cm}}\end{array}$

Therefore, the value of the displacement of the effort force is $3.14\text{cm}$ .

Conclusion:

Hence, all the changes quantities are evaluated.