Objects in equilibrium beam problem

[BrainMan]

Problem: A uniform beam of length 4 m and mass 10 kg, connected to the side of a building by a pivot hinge, supports a 20-kg light fixture,as shown in figure 4.43. Determine the tension in the wire and the components of the reaction force at the pivot.

Equations: sum of forces =0
Sum of torques=0

Attempt at problem: Since the sum of the torques must be zero the torque of the cable must be equal to the torque of the weights of both objects so T( 4 cos 45)= 20( 3 cos 45) + 98( cos 45). Where T is the tension in the rope. The correct answer is T= 160

 

[PhanthomJay]

You are in general not calculating torques correctly. Its force times perp distance or use the cross product rule for torques. Identify your pivot point and show your free body diagram .

 

[BrainMan]

 

[dauto]

Why are you using cos 45 when calculating the torque of the tension?

 

[BrainMan]

Because the distance of 4 is along the length of the beam and you must find the actual x distance from the pivot point by using cos 45.

 

[dauto]

Because the distance of 4 is along the length of the beam and you must find the actual x distance from the pivot point by using cos 45.

No you don't. Look again. there are two different angles in your picture.

 

[BrainMan]

I know the second angle is 60 degrees I just don't understand how to use it.

 

[dauto]

I know the second angle is 60 degrees I just don't understand how to use it.

Create an imaginary line connecting the pivot to the cable at 90° to the cable forming a triangle. use that triangle to find the distance used in the torque formula. The equation you used implies that triangle as an angle of 45 ° somewhere in it but that's not correct. It's a fairly simple trigonometry/geometry problem.

 

[BrainMan]

Can you send me a diagram? I'm having trouble visualizing the triangle.

 

[dauto]

Draw a line perpendicular to the cable making sure it passes over the pivot.

 

[BrainMan]

Like this?

 

[PhanthomJay]

View attachment 68745
Like this?

i don't see any perpendicular line . It would run from the side T to the pivot, at a right angle to T. The force T times that perp dist to pivot would give the torque caused by T about the pivot. Or try cross product rule it might be easier if you are at all familiar with it.
Incidentally, your torques for the applied weight forces are wrong also. In one case you used mass instead of weight, in the other , you forgot to include the moment arm length. Might be a careless error on those ?

 

[BrainMan]

i don't see any perpendicular line . It would run from the side T to the pivot, at a right angle to T. The force T times that perp dist to pivot would give the torque caused by T about the pivot. Or try cross product rule it might be easier if you are at all familiar with it.
Incidentally, your torques for the applied weight forces are wrong also. In one case you used mass instead of weight, in the other , you forgot to include the moment arm length. Might be a careless error on those ?

Should I have 98(2 cos 45) + 196(3 cos 45) for the applied weight? If so what makes the distance measurement for T different since it is on the same beam? I also have never heard of the cross product rule.

 

[BrainMan]

Originally I tried to break down T into its X Y components so T sin 60 (4 cos 45) + T cos 60 (4 cos 45) - 98(2 cos 45) - 196( 3 cos 45)= 0. Is this more near what I should be doing?

 

[PhanthomJay]

Should I have 98(2 cos 45) + 196(3 cos 45) for the applied weight? If so what makes the distance measurement for T different since it is on the same beam? I also have never heard of the cross product rule.

Yes that's correct for the applied weight torques. Note that torque is force times perpendicular distance . The perpendicular distance from the tension force to the pivot seems to be stumping you. It just requires some geometry and trig. Now you could break up T into its x and y components and sum the torques of each component which you tried to do, but you didn't calculate the components correctly because you used the wrong angle . I see the basic trig and geometry is troubling you. On the other hand, you handled the geometry and trig pretty good when doing the weight torques. Hmmm. Well' heck there are so many ways to calculate torque try the cross product T (r)(sin theta), where T is T , r is the distance from the point of application of T to the pivot ( the length of the beam), and theta is the angle between T and r. But better to practice up on your basic geometry first.

 

[dauto]

Originally I tried to break down T into its X Y components so T sin 60 (4 cos 45) + T cos 60 (4 cos 45) - 98(2 cos 45) - 196( 3 cos 45)= 0. Is this more near what I should be doing?

Actually for the first two terms you should have either

T sin 15° (4 cos 45°) + T cos 15° (4 sin 45°)

(because the angle between the cable and the horizontal is not 60°)

or

T sin 60° (4 cos 90°) + T cos 60° (4 sin 90°)

(If you decide to break the force into components parallel and perpendicular to the beam.

Also note that one of the cos 45° was replaced with the correct sin 45°.

But none of that is necessary and both of these expressions reduce (through simple trigonometry) to the expression that is more easily obtained by using the triangle I mentioned in earlier posts.

 

[BrainMan]

Thanks! I understand what I did wrong. And thanks to everyone who helped me with this problem. I thank you all for your great amount of patience!