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hangingwire
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Taking a physics 12 class in night school and he only gave us an example of a Beam and Wire question where the Wire was holding a Beam against a wall. The wire was at the end of the beam and holding a sign.
Using the methods taught, I have been able to understand the Torque = 0, but Forces get me stuck
This is a nifty site and my next class (and unit test) is on Tuesday. Teacher hardly checks emails!
Find the cord tension and force from hinge on this beam.
Hinge is at the wall. Beam is 4.50m long. Object at end is 20 kg heavy. The beam is 80kg. Also, the wire this time is 1.0m from the end (**star** is the wire) with an angle of 25 degrees.
H.______*__
Sum of T = 0
Sum of F = 0
First using the sum of Torques:
Tcw = TccwClockwise applications:
Mb*G (Beam)
Mo*G (Object)
Counter-cw:
Ty
Using all these applications and their distances relative to the pivot (Pivot @ H)
Mb*G*d + Mo*G*d = Ty*d
(80)(9.8)(2.25m) + (20)(9.8)(4.5m) = Tsin25 * (3.5m)
1764 + 882 = (3.5) * Tsin25
Divide by 3.5
756 = Tsin25
Divide by sin25
T = 1788.8N --> 1790N (The answer is rounded in my book)Now for forces.
Fx = 0
FHx - Tx = 0
FHx = Tx
FHx = Tcos25
FHx = 1790cos25
FHx = 1622.3N
But the answer is 1600N. Using pythagorean theorem, the x is larger than the resultant?
Using the methods taught, I have been able to understand the Torque = 0, but Forces get me stuck
This is a nifty site and my next class (and unit test) is on Tuesday. Teacher hardly checks emails!
Homework Statement
Find the cord tension and force from hinge on this beam.
Hinge is at the wall. Beam is 4.50m long. Object at end is 20 kg heavy. The beam is 80kg. Also, the wire this time is 1.0m from the end (**star** is the wire) with an angle of 25 degrees.
H.______*__
Homework Equations
Sum of T = 0
Sum of F = 0
The Attempt at a Solution
First using the sum of Torques:
Tcw = TccwClockwise applications:
Mb*G (Beam)
Mo*G (Object)
Counter-cw:
Ty
Using all these applications and their distances relative to the pivot (Pivot @ H)
Mb*G*d + Mo*G*d = Ty*d
(80)(9.8)(2.25m) + (20)(9.8)(4.5m) = Tsin25 * (3.5m)
1764 + 882 = (3.5) * Tsin25
Divide by 3.5
756 = Tsin25
Divide by sin25
T = 1788.8N --> 1790N (The answer is rounded in my book)Now for forces.
Fx = 0
FHx - Tx = 0
FHx = Tx
FHx = Tcos25
FHx = 1790cos25
FHx = 1622.3N
But the answer is 1600N. Using pythagorean theorem, the x is larger than the resultant?