Put a uniform meter scale horizontally on your extended index fingers with the left one at and the right one at . When you attempt to move both the fingers slowly towards the center, initially only the left finger slips with respect to the scale and the right finger does not. After some distance, the left finger stops and the right one starts slipping. Then the right finger stops at a distance from the center () of the scale and the left one starts slipping again. This happens because of the difference in the frictional forces on the two fingers. If the coefficients of static and dynamic friction between the fingers and the scale are and , respectively, the value of (in cm) is ______.
Answer:
Or read the worked solution below. Jump to the steps ↓
Written solution
Step 1: Define the initial coordinates and parameters.
The uniform meter scale has its center of mass at .
Step 2: Determine the normal forces on each finger.
Taking torque about the contact points of the left and right fingers yields the normal forces.
Step 3: Establish the condition for slipping.
The left finger slips while the right finger grips as long as the kinetic friction on the left is less than or equal to the maximum static friction on the right.
Step 4: Calculate the position where the left finger stops slipping.
The left finger slips, decreasing from while remains constant at , until the slipping condition reaches its limit.
Step 5: Calculate the position where the right finger stops slipping.
The right finger now slips, decreasing from while the left finger remains stationary at , until the roles reverse again.
The distance of the right finger from the center at this instant is .
Answer: 25.6