Question 8: Kitten on a Rod (Torque Balance)
Step 1: Force Equilibrium
Total downward force = Weight of rod + Weight of kitten = \( 10 + 20 = 30 \) N.
The rod is held by vertical friction at both ends: \( f_L + f_R = 30 \) N.
Constraint: Max friction is 20 N. So, \( f_L \le 20 \) and \( f_R \le 20 \).
This implies both \( f_L \) and \( f_R \) must be at least 10 N (e.g., if \( f_L=10 \), \( f_R=20 \)).
Step 2: Torque Equation
Let the kitten be at distance \( x \) cm from the left wall. Take torque about the right end.
Clockwise Torque (Lift): \( f_L \times 100 \)
Counter-Clockwise Torque (Load): \( 10 \times 50 \) (Rod) + \( 20 \times (100-x) \) (Kitten)
Step 3: Finding Safe Range
We apply the condition \( 10 \le f_L \le 20 \).
- Upper Bound: \( 25 – 0.2x \le 20 \implies 5 \le 0.2x \implies 25 \le x \)
- Lower Bound: \( 25 – 0.2x \ge 10 \implies 15 \ge 0.2x \implies 75 \ge x \)
Answer: Safe range is 25 cm to 75 cm from the left end.