1.

If a rod of mass 0.2 kg is in equilibria in given condition then friction force acting on rod in newtons is. [Force F is acting horizontally and g=10 m/s2] (Take √3=1.73). Write upto two digits after the decimal point.

Answer»
If a rod of mass 0.2 kg is in equilibria in given condition then friction force acting on rod in newtons is. [Force F is acting horizontally and g=10 m/s2] (Take 3=1.73). Write upto two digits after the decimal point.


Discussion

No Comment Found

Related InterviewSolutions