Stuck on Torque Question (W/ Picture)

[tachu101]

Homework Statement

A uniform beam of 6.6 meters with a weight of 42.67 Newtons is pinned to a wall and supports a weight of 19.62 Newtons.The beam is angled up at 16 degrees and the right end of the beam is supported by a cable attached to the other side of the wall. Find the value of; Tension of the cable, horizontal Pin Joint Force, vertical Pin Joint Force, and the resultant of the H and V force components (where the angle is measured counter clockwise from the horizontal).

Homework Equations

I am not sure how to start this problem I only know that summing the forces in the x, y and torque direction will be used.

The Attempt at a Solution

I tried summing the forces

\sumx = H-T=0 ---- so H=T ? (but what are the numbers)

\sumy = V - 42.67Newtons(weight of beam) - 19.62Newtons(weight at end) = 0 (am I missing any?) This would give me V= 62.29 Newtons

(If pin joint is the pivot point)
\sum\tau = H(0)+V(0)-42.67N(3.3cos16)-19.62N(6.6cos16)+T=0
I don't know if this one is right, something does not seem right with the tension part.
If this is right that would make the Tension = 259.8N ; which seems too much

And because T=H I would think that H would also equal 259.8 Newtons


The last part I have no idea what to do.

 

[Doc Al]

\sumx = H-T=0 ---- so H=T ? (but what are the numbers)

Good. Don't worry about the numbers. You'll solve for the numbers by combining all three equations.

\sumy = V - 42.67Newtons(weight of beam) - 19.62Newtons(weight at end) = 0 (am I missing any?) This would give me V= 62.29 Newtons

Good.

(If pin joint is the pivot point)
\sum\tau = H(0)+V(0)-42.67N(3.3cos16)-19.62N(6.6cos16)+T=0
I don't know if this one is right, something does not seem right with the tension part.

You forgot the moment arm for the tension force. You need the torque due to the tension force, not just the force.

 

[tachu101]

So it would be

H(0)+V(0)-42.67N(3.3cos16)-19.62N(6.6cos16)+T(6.6cos16) or T(6.6sin16)=0

 

[Doc Al]

Well, which is it? Which gives the perpendicular distance to the line of force? The tension acts horizontally.

 

[tachu101]

um, if this was a test I probably would have to go with cos , but I would not be sure.

 

[tachu101]

H(0)+V(0)-42.67N(3.3cos16)-19.62N(6.6cos16)+T(6.6cos16)=0

T= 40.955 Newtons

 

[Doc Al]

um, if this was a test I probably would have to go with cos , but I would not be sure.

But you used cosine for the other forces, which were vertical. What's the difference?

Perhaps you should review calculating torques: http://hyperphysics.phy-astr.gsu.edu/hbase/torq2.html#tc"

 

[tachu101]

I think that I get it now so (vertical cosine / horizontal sine) ----

H(0)+V(0)-42.67N(3.3cos16)-19.62N(6.6cos16)+T(6.6sin16)=0

T= 142.8 Newtons

How do I start the last part?

 

[Doc Al]

How do I start the last part?

What's the last part? Finding the resultant?

You found T, H, and V. Now find the resultant of H and V. (Remember that these are just horizontal and vertical components of some resultant force. Use right triangle geometry to solve for the resultant and the angle it makes.)

 

[tachu101]

I this like adding vectors?

 

[tachu101]

I am getting 155.89 Newtons

 

[Doc Al]

I this like adding vectors?

Absolutely.

I am getting 155.89 Newtons

That looks about right.

 

[tachu101]

thanks so much

 

[Doc Al]

Don't forget to figure out what angle it makes. (I think they want that specified, not just the magnitude.)