Asked by Criticos
                A uniform metre rule of mass 150g  is pivoted freely at 0 cm mark.What force applied vertically upwards at the 60 cm mark is needed to maintain the rule horizontally?
            
            
        Answers
                    Answered by
            R_scott
            
    this is a class 2 lever
the weight of the rule acts at its center of mass
f * 60 = (.15 g) * 50 ... g is gravitational acceleration ... 9.8 m/s^2
    
the weight of the rule acts at its center of mass
f * 60 = (.15 g) * 50 ... g is gravitational acceleration ... 9.8 m/s^2
                    Answered by
            bett
            
    which is the answer
    
                    Answered by
            bett
            
    show the workings
    
                    Answered by
            aron smash
            
    i din't get answer
    
                    Answered by
            rotich 
            
    how to find 
    
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