(10%) Problem 6: Two children push on opposite sides of a door during play. Both

Question

(10%) Problem 6: Two children push on opposite sides of a door during play. Both push horizontally and perpendicular to the surface of the door. One child pushes with a force of 16 N at a distance of 0.595 m from the hinges, and the second child pushes at a distance of 0.44 m C What is the magnitude of the force, in newtons, the second child must exert to keep the door from moving? Assume friction is negligible. Grade Summary Deductions Potential 0% 100% ?? sin() cos() tan() ? () 789 cotanasin)acosO atan acotan) sinh) coshO tanhOcotanh0 1 2 3 0 BACKSPA DEL Submissions Attempts remaining:4 (5% per attempt) detailed view o Degrees O Radians 0 Submit Hint Hints: 1 % deduction per hint. Hints remaining: 2 Feedback: 1% deduction per feedback. All content © 2018 Expert TA, LLC

Explanation / Answer

To stop the door from moving, Net torque must be zero

We know that

Torque = Force*distance

So,

Net torque = 0

Clockwise Torque - Counter-Clockwise Torque = 0

Clockwise Torque = Counter-Clockwise Torque

F1*r1 = F2*r2

F2 = F1*(r1/r2)

Using given values:

F2 = 16*(0.595/0.44)

F2 = 21.64 N

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