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 3.5-kN load.(adsbygoogle = window.adsbygoogle || []).push({});

I have attached an image of the problem.

1. The problem statement, all variables and given/known data

2. Relevant equations

ƩM_{O}= 0

ƩF_{x}=0

ƩF_{y}=0

3. 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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# Homework Help: Find Magnittude of the Force Supported by the bearing O

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