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Two children push on opposite sides of a door during play. Both push horizontally and perpendicular to the door. One child pushes with a force of 17.5 N at a distance of 0.600 m from the hinges, and the second child pushes at a distance of 0.450 m. What force must the second child exert to keep the door from moving

Respuesta :

Answer:

Explanation:

We shall apply principle of moment to solve the problem .

For rotational balance , moment of force created by both the children will be equal and opposite .

moment created by one child = 17.5 x .6 = 10.5 Nm

moment of force created by other child = F x .45 Nm where F is required force

F x .45 = 10.5

F = 23.33 N .