# Find Magnittude of the Force Supported by the bearing O

Northbysouth
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.

I have attached an image of the problem.

ƩMO = 0
ƩFx =0
ƩFy =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
ƩMO = 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 (Rx) and a force in the y direction (Ry) I calculated these:

ƩFx= 0
0 = -Rx + Tcos(24.5196)
Rx = 4.899 kN

ƩFy = 0
0 = -3.5 kN + Ry -Tsin(24.5196)
Ry = 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.

#### Attachments

• fastened pulleys R.png
29.1 KB · Views: 1,502

Homework Helper
Gold Member
It's a significant figure error....too many places after the decimal point...try 7.5 kN

Northbysouth
7.5 kN didn't work either. I forget to include this in the image but the system has a tolerance to +/- 1 to the third significant digit.

Northbysouth
I found my mistake, the angle that I calculated using arctan(130/285) was wrong. It should have been arcsin(130/285) because 285mm is no the longest length of the triangle.

arcsin(130/285) = 27.138

Substituting 27.138 into my calculations I get fx = 4.7918kN and fy= 5.9561kN

Taking the magnitude of these gives me a resultant of 7.644 kN which is the correct answer.