Max Distance Mass M Can Be Placed Before Wire Breaks

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SUMMARY

The maximum distance from the wall at which mass M can be placed before the wire breaks is 1.5 meters. This conclusion is derived from analyzing the forces acting on a uniform horizontal bar of length 5 meters and weight 233 N, supported by a wire at an angle of 43 degrees. The wire can withstand a maximum tension of 547 N, which is calculated to be 731 N when considering the horizontal component. The equilibrium condition shows that any distance greater than 1.5 meters will exceed the wire's tension capacity, leading to failure.

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A uniform horizontal bar of length L = 5 m and weight 233 N is pinned to a vertical wall and supported by a thin wire that makes an angle of theta = 43o with the horizontal. A mass M, with a weight of 384 N, can be moved anywhere along the bar. The wire can withstand a maximum tension of 547 N. What is the maximum possible distance from the wall at which mass M can be placed before the wire breaks?
 
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:: Use Equilibrium condition ::
 


To determine the maximum distance from the wall at which mass M can be placed before the wire breaks, we need to consider the forces acting on the system. The weight of the bar and the mass M will create a downward force, while the tension in the wire will create an upward force.

First, we need to find the tension in the wire at the maximum distance. We can use trigonometry to determine that the horizontal component of the tension is Tcos(theta), where theta is the angle between the wire and the horizontal. Plugging in the given values, we get Tcos(43) = 547 N. Solving for T, we get T = 547 N / cos(43) = 731 N.

Next, we need to find the distance from the wall at which the tension in the wire is equal to its maximum capacity of 547 N. This can be done by setting up an equation that equates the tension in the wire to the weight of the bar and the mass M. We know that the weight of the bar is 233 N, and the weight of the mass M is 384 N. So, the equation becomes:

T = 233 N + 384 N

Substituting the value of T we found earlier, we get:

731 N = 233 N + 384 N

Solving for the distance, we get:

d = (731 N - 233 N) / 384 N = 1.5 m

Therefore, the maximum possible distance from the wall at which mass M can be placed before the wire breaks is 1.5 m. Any distance greater than this will result in a tension in the wire that exceeds its maximum capacity of 547 N, causing it to break. It is important to note that this maximum distance is only valid if the bar and the wire are ideal and there is no friction or other external forces acting on the system.
 

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