Chapter 9, Problem 15P

The force required to pull the cork out of the top of a wine bottle is in the range of 200 to 400 N.What range of forces F is required to open a wine bottle with the bottle opener shown in Fig. 9-55?

Figure 9-55

To determine

The range of forces is required to open a wine bottle with the bottle opener shown in figure.

Answer to Problem 15P

Solution: $22.78\text{}\text{N}\text{to 54}\text{.6 N}$

Explanation of Solution

Given info:

The force required to pull the cork out of the top of a wine bottle is in the range of 200 N to 400N.

Formula used:

  $\text{The net torque}={\sum }_i{r}_i\times F_i$

Here, $F_i$ is the applied force and $r_i$ represents its corresponding distance from the rotational axis.

Calculation:

The force Fapplied at a distance of 79 mm from the pivot (point pushing on the bottle) and to pull up effective weight in the range of 200 to 400 N a distance of 9 mm from the pivot point.

Therefore, the minimum force applied will be when the net torque is zero.

torque $={\sum }_i{r}_i\times F$

This leads to (choosing counter clockwise about pivot as positive torque)

  $\begin{array}{l}\left|F_1\right|\left(79\text{mm}\right)-\left(200\text{N}\right)\left(9\text{mm}\right)=0\\ \left|F_1\right|=\frac{200\times 9}{79}=22.78\text{N}\\ \left|F_2\right|\left(79\text{mm}\right)-\left(400\text{N}\right)\left(9\text{mm}\right)=0\\ \left|F_2\right|=\frac{400\times 9}{79}=54.6\text{N}\end{array}$

Conclusion: The range of force is $22.78\text{}\text{N}\text{to 54}\text{.6 N}$ .