Problem 11.50 End A of the bar AB in rests on a frictionless horizontal surface,
ID: 1437197 • Letter: P
Question
Problem 11.50 End A of the bar AB in rests on a frictionless horizontal surface, and end B is hinged. A horizontal force F of magnitude 290 Nis exerted on end A. lg the weight of the bar. nore 5.00 m 4.00 m Part A What is the horizontal component of the force exerted by the bar on the hinge at B? Express your answer with the appropriate units. Value Units Submit My Answers Give Up Part B What is the vertical component of the force exerted by the bar on the hinge at B? Express your answer with the appropriate units. F. Value Units Submit My Answers Give UpExplanation / Answer
Applying Newton’s Second Law along the horizontal and vertical
directions,
samesion Fx = max = F - Fh
= Fh = F = - 290N
samesion Fx = may = Fn - Fv = 0
Applying the Second Law for Rotation about the origin,
In the same manner, you will also find by summing vertical forces that the vertical reaction force at the hinge must be equal and opposite to the vertical reaction force on the floor.
If you sum moments about the hinge, you will be able to figure the vertical reaction force Fv
I'll assume that CCW is positive
given the other assumptions, the only forces tending to create a moment are the horizontal applied force of 160 N and the reaction force on the floor Fv
Each moment is a force multiplied by it's lever arm
from pythagoras we find we have a 3-4-5 triangle
samestion = I alpha => 4Fh - 3Fh => 4Fh => 3Fv
Fv = -F * 4 / 3
Fv = -290 * 4 / 3
Fv = - 386.66