91Ó°ÊÓ

When a cork is opened, there is no cork rotation; the only cork moves upward.The cork does not rotate, and the net torque on the cork is zero due to the force on the opener and the pull force.

The required force to open the cork is \({F_ \circ } = 200\;{\rm{N}}\;{\rm{to}}\;400\;{\rm{N}}\).

The force on the opener is F.

According to Newton’s third law, the reaction force on the cork is downward. Here, the pivot point is at the edge where the end of the opener is touched with the bottle.

The free-body diagram of the problem is shown below.

Now, the condition for no rotation of the torque is

\(\begin{array}{c}F \times \left( {70\;{\rm{mm}} + 9\;{\rm{mm}}} \right) - {F_ \circ } \times \left( {9\;{\rm{mm}}} \right) = 0\\F = \frac{9}{{79}}{F_ \circ }\end{array}\).

Now, substituting the value of \({F_ \circ }\) in the above equation,

\(\begin{array}{c}F = \frac{9}{{79}}\left( {200\;{\rm{N}}\;{\rm{to}}\;400\;{\rm{N}}} \right)\\F = 22.78\;{\rm{N}}\;{\rm{to}}\;45.57\;{\rm{N}}\\F \approx 20\;{\rm{N}}\;{\rm{to}}\;50\;{\rm{N}}\end{array}\).

Hence, the required range of force on the opener is 20 N to 50 N.