Chapter 3, Problem 1MDP

1. 1. Prove the law of the lever.

To determine

Prove the law of the lever.

Answer to Problem 1MDP

The law of the lever is proved.

Explanation of Solution

Consider the lever consists of rigid bar is shown in Figure 1.

According to the law of lever, in equilibrium condition the torque due to effort force is equal to the torque load.

${\tau }_{\text{effort}}={\tau }_{\text{load}}$ (1)

Refer to Figure 1, write the formula for ${\tau }_{\text{effort}}$ and ${\tau }_{\text{load}}$.

${\tau }_{\text{load}}=F_{\text{e}}{d}_{\text{e}}$ (2)

${\tau }_{\text{load}}=F_{\text{l}}{d}_{\text{l}}$ (3)

Substitute equation (2) and (3) in equation (1)

$F_e{d}_e=F_{\text{l}}{d}_{\text{l}}$ (4)

Proof:

Consider the uniform bar ABCD suspended by middle point M as shown in Figure 2.

Refer to Figure 2; the bar is divided into two parts as AEFD and EBCF. Consider the middle point of AEFD is K and the middle point of EBCF is L.

The force of AEFD is represented as $F_{\text{AEFD}}$ and the force of EBCF is represented as $F_{\text{EBCF}}$. The force $F_{\text{AEFD}}$ and $F_{\text{EBCF}}$ are proportional to GI and IH respectively.

Write the relation between two forces.

$\frac{F_{\text{AEFD}}}{F_{\text{EBCF}}}=\frac{GI}{IH}$ (5)

Consider the forces $F_{\text{AEFD}}$ and $F_{\text{EBCF}}$ are concentrated at K and L.

Modify equation (4) for $F_{\text{AEFD}}$ and $F_{\text{EBCF}}$ as follows.

$F_{\text{AEFD}}\times MK=F_{\text{EBCF}}\times ML$

$\frac{F_{\text{AEFD}}}{F_{\text{EBCF}}}=\frac{ML}{MK}$ (6)

Substitute $\frac{ML}{MK}$ for $\frac{F_{\text{AEFD}}}{F_{\text{EBCF}}}$ in equation (5) to prove the law of the lever.

$\frac{GI}{IH}=\frac{ML}{MK}$ (7)

Consider $l$ and $b$ is used to name the GH bar and MI bar respectively.

$\begin{array}{l}GH=l\\ MI=b\end{array}$

Write the expression for GI.

$GI=\frac{1}{2}GH+MI$

Substitute $l$ for GH and $b$ for MI to find GI.

$\begin{array}{c}GI=\frac{l}{2}+b\\ =\frac{l+2b}{2}\end{array}$

Write the expression for IH.

$IH=GH-GI$

Substitute  $l$ for GH and $\frac{l+2b}{2}$ for GI to find IH.

$\begin{array}{c}IH=l-\left(\frac{l+2b}{2}\right)\\ =\frac{2l-l-2b}{2}\\ =\frac{l-2b}{2}\end{array}$

Write the expression for IH.

$ML=MI+\frac{IH}{2}$

Substitute $b$ for MI and $\frac{l-2b}{2}$ for IH to find ML.

$\begin{array}{c}ML=b+\frac{\left(\frac{l-2b}{2}\right)}{2}\\ =b+\frac{l-2b}{4}\\ =\frac{4b+l-2b}{4}\\ =\frac{l+2b}{4}\end{array}$

Write the expression for GK.

$GK=\frac{GI}{2}$

Substitute $\frac{l+2b}{2}$ for GI to find GK.

$\begin{array}{c}GK=\frac{\left(\frac{l+2b}{2}\right)}{2}\\ =\frac{l+2b}{4}\end{array}$

Write the expression for MK.

$MK=GM-GK$

Substitute $\frac{1}{2}GH$ for GM.

$MK=\frac{1}{2}GH-GK$

Substitute  $l$ for GH and $\frac{l+2b}{4}$ for GK to find MK.

$\begin{array}{c}MK=\frac{l}{2}-\frac{l+2b}{4}\\ =\frac{2l-l-2b}{4}\\ =\frac{l-2b}{4}\end{array}$

Substitute $\frac{l+2b}{2}$ for $GI$ and $\frac{l-2b}{2}$ for $IH$ to find the ratio of $GI\text{to}IH$.

$\frac{GI}{IH}=\frac{\left(\frac{l+2b}{2}\right)}{\left(\frac{l-2b}{2}\right)}$

$\frac{GI}{IH}=\frac{l+2b}{l-2b}$ (8)

Substitute $\frac{l+2b}{4}$ for $ML$, and $\frac{l-2b}{4}$ for $MK$ to find the ratio of $ML\text{to}MK$.

$\frac{ML}{MK}=\frac{\left(\frac{l+2b}{4}\right)}{\left(\frac{l-2b}{4}\right)}$

$\frac{ML}{MK}=\frac{l+2b}{l-2b}$ (9)

Compare equations (8) and (9).

$\frac{GI}{IH}=\frac{ML}{MK}$ (10)

Compare equation (7) and (10), which satisfies the law of the lever.

Conclusion:

Thus, the law of the lever is proved.