A solid cone of mass (m = 1 kg) is placed on a plank of mass M = 1 kg which is placed on a smooth horizontal plane. The coefficient of friction between cone and the plank is 1/4. If a horizontal force F is applied on the plank, then find the maximum value of F for which the cone is in equilibrium with respect to plank. (Take g = 10 m/s² and H/r = 20)
Physics · Mixed Chapter
JEE Mains

Question

A solid cone of mass (m = 1 kg) is placed on a plank of mass M = 1 kg which is placed on a smooth horizontal plane. The coefficient of friction between cone and the plank is 1/4. If a horizontal force F is applied on the plank, then find the maximum value of F for which the cone is in equilibrium with respect to plank. (Take g = 10 m/s² and H/r = 20)

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: If cone and plank move together, their horizontal acceleration is a=F/(M+m)=F/2. Step 2: Friction limits acceleration to a≤μg=(1/4)(10)=2.5 m/s². Step 3: The cone can topple when the resultant of mg and pseudo-force through its centre of mass passes beyond its circular base. For a uniform solid cone its centre of mass is at height H/4. Step 4: At the tipping threshold a/g = r/(H/4)=4r/H=4/20=1/5, so aₘₐₓ=2 m/s², more restrictive than the friction limit. Step 5: Fₘₐₓ=(M+m)aₘₐₓ=2×2=4 N.