: A door is made of a 29-kg uniform board that measures 2.3 m high and 0.979 m w
ID: 1776232 • Letter: #
Question
:
A door is made of a 29-kg uniform board that measures 2.3 m high and 0.979 m wide. The door is free to rotate about its hinges. A force of magnitude 2.32 N is applied to the door.
(a) What is the magnitude of the maximum torque that can be generated about the door's hinges?
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2.27 N·m
(b) With the torque calculated in part (a), what is the magnitude of the angular acceleration of the door?
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.245 rad/s2
(c) Assuming that the torque calculated in part (a) is constant, what is the rotational power being supplied to the door after 2.87 s? Assume the door starts from rest.
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??? W
___________________________________________________________________________________________________________________________________________
Enter your answer
52.009 N
(b) What is the vertical component of the force exerted by the hinge on the rod?
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???? N
(c) What is the horizontal component of the force exerted by the hinge on the rod?
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45.404 N
Explanation / Answer
As per guide lines I work first question please post remaining question in the next post
(a)
formula for maximum torque is
T = rF
=0.979 ( 2.32)
=2.27 Nm
(b)
T = I alpha
alpha = T/ I
= T/mw^2/3
= 2.27128/29 ( 0.979)^2/3
=0.245 rad/s^2
(3)
formula for rotational power is
P = Twf
= T(wo+ aplat t)
= 2.27128( 0+0.245( 2.87)
=1.598 W
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