Static Equilibrium of mass and wire

In summary, the problem involves a 345 kg mass supported by a wire attached to a 15 m long steel bar that is pivoted at a vertical wall and supported by a cable. The mass of the bar is 95 kg. Using the given information and equations, the tension in the cable and the force exerted by the wall on the steel bar can be calculated. For part (a), the tension in the cable is found to be 10003.6976 N and the force exerted by the wall on the steel bar is 5001.8488 N in the i direction and -4347.056 N in the j direction. For part (b), with a longer cable attached to the upper end of the
  • #1
chara76
10
0

Homework Statement



A 345 kg mass is supported on a wire attached to a 15 m long steel bar that is pivoted at a vertical wall and supported by a cable. The mass of the bar is 95 kg. (Take right and up to be positive.)

http://img522.imageshack.us/img522/8374/1265alt.gif" [Broken]

(a) With the cable attached to the bar 5.0 m from the lower end as shown, find the tension in the cable.
10003.6976 N (Done)

Find the force exerted by the wall on the steel bar.
5001.8488 N i + -4347.056 N j (Done)

(b) A somewhat longer cable replaces the old and is attached to the steel bar 5.0 m from its upper end, connecting to the same place on the wall as before and maintaining the same angle between the bar and the wall. Find the tension in the cable.
? N

Find the force exerted by the wall on the steel bar.
? N i + ? N j

Homework Equations



Sum of Torques = 0
Sum of Fy = 0
Sum of Fx = 0

The Attempt at a Solution



(b)
[(95kg)(9.81)(7.5m)(cos60) + (345kg)(9.81)(15m)(cos60)] / 10m
= 2887 N This is wrong

I haven't attepmted the second part of B because I need the tension to do so.
 
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  • #2
unable to portrait the picture of problm..
can you please put up the figure..
 
  • #3
Added link to picture. Sorry about that
 
  • #4
it shud be:
[(95kg)(9.81)(7.5m)(sin60) + (345kg)(9.81)(15m)(sin60)] / (10cos30)
 
  • #5
I have this same problem, but I already got part a. Can someone please help me find part b?
 
  • #6
I found the force, and the force in the i direction, but not in the j...
 

What is static equilibrium of mass and wire?

Static equilibrium of mass and wire refers to a state in which the mass and wire are at rest and there is no net force acting on them. This means that the forces acting on the mass and wire are balanced and there is no acceleration in any direction.

How is static equilibrium achieved in a mass and wire system?

Static equilibrium is achieved in a mass and wire system when the downward force of the mass due to gravity is balanced by the upward force of the wire. This is also known as the tension force, which pulls the wire upwards to counteract the weight of the mass.

What factors affect static equilibrium in a mass and wire system?

The factors that affect static equilibrium in a mass and wire system include the weight of the mass, the length of the wire, and the angle at which the wire is suspended. Any changes in these factors can alter the tension force and disrupt the static equilibrium.

What are some real-world applications of static equilibrium in a mass and wire system?

Static equilibrium in a mass and wire system is commonly used in construction and engineering to support structures such as bridges and cranes. It is also used in physics experiments to study the properties of tension forces and their effects on objects in equilibrium.

How can we calculate the tension force in a mass and wire system?

The tension force in a mass and wire system can be calculated using the formula T = mg, where T is the tension force, m is the mass of the object, and g is the acceleration due to gravity. This formula assumes that the wire is massless and inextensible.

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