- #1

Northbysouth

- 249

- 2

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.

ƩM

ƩF

ƩF

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

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

ƩF

0 = -R

R

ƩF

0 = -3.5 kN + R

R

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}= 0ƩF

_{x}=0ƩF

_{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}= 00 = (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}= 00 = -R

_{x}+ Tcos(24.5196)R

_{x}= 4.899 kNƩF

_{y}= 00 = -3.5 kN + R

_{y}-Tsin(24.5196)R

_{y}= 5.7305 kNThus 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.