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3 years ago

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Two waiters are trying to get through a single door of a kitchen. One pushes on one side of a door 0.567 m from the hinge with a

force of 257 N perpendicular to the door, the other pushes from the other side of the door 0.529 m from the hinge. The door doesn't move. With what perpendicular force is the second waiter pushing on the door?

1 answer:

NISA [10]3 years ago

7 0

Answer:

$275.5 N$

Explanation:

$F_{1}$ = Force on one side of the door by first waiter = 257 N

$F_{2}$ = Force on other side of the door by second waiter

$r_{1}$ = distance of first force by first waiter from hinge = 0.567 m

$r_{2}$ = distance of second force by second waiter from hinge = 0.529 m

Since the door does not move. hence the door is in equilibrium

Using equilibrium of torque by force applied by each waiter

$r_{1} F_{1} = r_{2} F_{2} \\(0.567) (257) = (0.529) F_{2}\\F_{2} = 275.5 N$

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