Calculating Horizontal Force for Displaced Hanging Object

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To calculate the horizontal force needed to hold a 125 kg mail bag displaced 1.9 m sideways while suspended from a 3.9 m rope, one must consider the forces acting on the bag. The problem involves resolving the tension in the rope into horizontal and vertical components, necessitating a free body diagram. The gravitational force acts downward, while the tension creates an angle with the vertical, influencing the required lateral force. Understanding the difference between mgsin and mgcos is crucial for accurate calculations. This scenario is a statics problem that requires careful consideration of all forces involved.
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A 125kg mail bag hangs by a vertical rope 3.9 m long. A postal worker then displaces the bag to a position 1.9 m sideways from its original position, always keeping the rope taut.

What horizontal force is necessary to hold the bag in the new position?



I used the gravitational potential energy formula: mgy1-mgy2 and got 2450 but that is not the right answer. And I do not understand what they mean by moving it sidways. Is that why I am getting the answer wrong?
 
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The mass of 125 kg suspended from the 3.9 m rope (which remains taut) is displaced sideways (horizontally) by 1.9 m. The mass must swing a circular arc of radius 3.9 m.

Make a diagram of a triangle. The rope forms the hypotenuse and the 1.9 m displacement is the base. This is a statics problem and one must find the lateral force required to keep the 125 kg mass at rest.

The weight acts downward (vertically) with gravity, and the tension of the rope acts at some angle with respect to horizontal (or vertical). Resolve the tension of the rope into horizontal and vertical components.
 
first thing to do is draw a free body diagram. Make sure to include ALL the forces. this mgsin, mgcos, Normal force, and gravity. To help you answer your question, know the difference between mgsin and mgcos.
 
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