# X and Y - components of a force

1. Oct 11, 2008

### Godblessyou

1. The problem statement, all variables and given/known data
Calculate the x- and y- components of the force C which member BC exerts on
member ACD. The cables are wrapped securely around the two pulleys, which are
fastened together.
(Note the different diameters of the pulleys)
Problem

2. Relevant equations

The sum of moment in the systems = 0
The sum of force in Fx and Fy direction in an equilibrium condition equals zero.

Ok what I did was I differentiated each component and tried to calculate, but it didn't work. IN the attached file the question is shown.

Please Please I beg you all for help I have wasted a whole three days on it and even today I have slept for only 4 hours. Please it is due tommorow please help me.

#### Attached Files:

• ###### Statics assignment 3.pdf
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2. Oct 12, 2008

### Q_Goest

Re: Force!1

I'm assuming the duel pully set up pivots around point D. So your 100 kg mass creates a force downwards and a torque (moment) around D at a distance (radius) of 0.5 m which is resisted ONLY by the second cable with a radius of .25 m. From that information alone, you can determine the tension in the second cable since it's simply the ratio of the radius.

Now you have the tension in the second cable (from wall to pully) which creates a force in the X direction (horizontal) and a force down from the 100 kg weight in the Y direction (verticle). You also have two forces created by the beams at 45 degree to the wall. The sum of all the forces in the Y is the weight down and the verticle component of each beam at the wall. Similarly, the sum of all forces in the X direction is the second cable force and the horizontal component of each beam at the wall.

You also know that for each of the beams (ex: BC) there are no moments at the pinned locations so the force is directed along the axis of the beam (ex: so for BC, the X component of the force is equal to the Y component).

Write down the two equations (sum Fx and sum Fy) and solve.