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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] |
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