Asked by Angela
A 3kg bar is laying on a smooth vertical wall of 3L of height and a floor of 2L. Determine the force that the wall exerts over the bar.
I have tried to get to length of the bar by using Pitagoras but no, then I tried using a free body diagram but I don't know the direction of the forces of the wall and floor.
Can someone help me?
I have tried to get to length of the bar by using Pitagoras but no, then I tried using a free body diagram but I don't know the direction of the forces of the wall and floor.
Can someone help me?
Answers
Answered by
Damon
well the length is L sqrt 13
but who cares?
force up from floor = m g = 3g
(because wall is smooth none up there)
force sideways from wall = F
thus force sideways at floor = F toward wall
Now moments about center
clockwise mg * 2L/2 = m g L
counter clockwise F*1.5 L + F*1.5 L
= 3 F L
so
m g L = 3 F L
F = m g /3 = 3 kg * 9.81 m/s^2 /3
= 9.81 Newtons
but who cares?
force up from floor = m g = 3g
(because wall is smooth none up there)
force sideways from wall = F
thus force sideways at floor = F toward wall
Now moments about center
clockwise mg * 2L/2 = m g L
counter clockwise F*1.5 L + F*1.5 L
= 3 F L
so
m g L = 3 F L
F = m g /3 = 3 kg * 9.81 m/s^2 /3
= 9.81 Newtons
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