Asked by Megan
                A school yard teeter totter with a total length of 5.46 m and a mass of 35.9 kg is pivoted at its center. A 17.7 kg child sits on one end of the teeter totter. Where should a parent push with a force of 272 N in order to hold the teeter totter level? 
            
            
        Answers
                    Answered by
            bobpursley
            
    sum moments about the center.
5.46/2 * 17.7g-272N*d=0 the sum of moments about any point equals zero. d is the distance away from the child, on the opposide side.
    
5.46/2 * 17.7g-272N*d=0 the sum of moments about any point equals zero. d is the distance away from the child, on the opposide side.
                    Answered by
            drwls
            
    Since the board is balanced at the center, the weight of the board does not contribute a moment.
You must balance the moment (force x distance) applied by the parent with the moment due to the weight of the child.
272*X = (17.7 kg)*g*(2.73 m)
Solve for X in meters
Note that the child's mass has to be multiplied by g to get the weight force in Newtons.
    
You must balance the moment (force x distance) applied by the parent with the moment due to the weight of the child.
272*X = (17.7 kg)*g*(2.73 m)
Solve for X in meters
Note that the child's mass has to be multiplied by g to get the weight force in Newtons.
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