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The two light pulleys are fastened together and form an integral unit. They are prevented from turning about their bearing at O by a cable wound securely around the smaller pulley and fastened to point A. Calculate the magnitude R of the force supported by the bearing O for the applied 3.5kN load.
I have attached an image of the problem.
ƩM_{O} = 0
ƩF_{x} =0
ƩF_{y} =0
First I found the angle between the positive x axis and the cable (AB)
θ = arctan(130/285)
θ = 24.5196°
Then I found the moment about point to calculate the tension (T) in the cable
ƩM_{O} = 0
0 = (3.5 kN)(200mm) (130mm)T
T = 5.3846 kN
Then, knowing that the pivot point, O, has a force in the x direction (R_{x}) and a force in the y direction (R_{y}) I calculated these:
ƩF_{x}= 0
0 = R_{x} + Tcos(24.5196)
R_{x} = 4.899 kN
ƩF_{y} = 0
0 = 3.5 kN + R_{y} Tsin(24.5196)
R_{y} = 5.7305 kN
Thus the magnitude R is:
R = sqrt((4.899 kN)^2 +(5.7305 kN)^2)
R = 7.53914 kN
It says my answer is wrong and I can't see where my mistake is. Help would be appreciated.
I have attached an image of the problem.
Homework Statement
Homework Equations
ƩM_{O} = 0
ƩF_{x} =0
ƩF_{y} =0
The Attempt at a Solution
First I found the angle between the positive x axis and the cable (AB)
θ = arctan(130/285)
θ = 24.5196°
Then I found the moment about point to calculate the tension (T) in the cable
ƩM_{O} = 0
0 = (3.5 kN)(200mm) (130mm)T
T = 5.3846 kN
Then, knowing that the pivot point, O, has a force in the x direction (R_{x}) and a force in the y direction (R_{y}) I calculated these:
ƩF_{x}= 0
0 = R_{x} + Tcos(24.5196)
R_{x} = 4.899 kN
ƩF_{y} = 0
0 = 3.5 kN + R_{y} Tsin(24.5196)
R_{y} = 5.7305 kN
Thus the magnitude R is:
R = sqrt((4.899 kN)^2 +(5.7305 kN)^2)
R = 7.53914 kN
It says my answer is wrong and I can't see where my mistake is. Help would be appreciated.
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