Tension in Strings Holding Up Shelf

In summary: My mistake was in thinking that the center of mass was at the point where the wires were attached to the shelf. In fact, the center of mass is at the center of the tool. The tension in the right hand wire is then 185.2040 kg cm-2 (8.469 kgm-2).
  • #1
ARL91
2
0
1. A 50.0 cm, uniform, 57.0 N shelf is supported horizontally by two vertical wires attached to the sloping ceiling (the figure ). A very small 26.0 N tool is placed on the shelf midway between the points where the wires are attached to it.
http://session.masteringphysics.com/problemAsset/1260552/1/YF-11-23.jpg

2. All I know is that The downward force which is 83 N (26 N+ 57 N) and upward force must be equal to zero because this is an equilibrium problem.
3. I have tried to find the center of gravity:
x=(m1x1+m2x2)/(m1+m2)--> 185.2040Kgcm/8.469 Kg = 21.867 cm from the left, after this I am unsure how to proceed.
 
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  • #2
ARL91 said:
1. A 50.0 cm, uniform, 57.0 N shelf is supported horizontally by two vertical wires attached to the sloping ceiling (the figure ). A very small 26.0 N tool is placed on the shelf midway between the points where the wires are attached to it.
http://session.masteringphysics.com/problemAsset/1260552/1/YF-11-23.jpg




2. All I know is that The downward force which is 83 N (26 N+ 57 N) and upward force must be equal to zero because this is an equilibrium problem.



3. I have tried to find the center of gravity:
x=(m1x1+m2x2)/(m1+m2)--> 185.2040Kgcm/8.469 Kg = 21.867 cm from the left, after this I am unsure how to proceed.

You can proceed by taking moments about the points where the shelf is supported.

Suppose you start with the leftmost support point. Assuming that all the mass of the shelf and tool are concentrated at their center of mass, what torque does this weight produce about the left hand suspension point? Then, if the torque caused by the tension in the right hand suspension wire must exactly counteract this (sum of the torques must be zero for a static system), what must the tension be?
 
  • #3
Thank you so much! Turns out I was using the wrong distance from the left wire.
 

Related to Tension in Strings Holding Up Shelf

1. What is tension in strings holding up a shelf?

Tension in strings holding up a shelf refers to the force applied by the strings to support the weight of the shelf. This tension force is necessary to keep the shelf in place and prevent it from falling.

2. How is tension in strings calculated?

Tension in strings can be calculated using the formula T = mg, where T is the tension force, m is the mass of the shelf, and g is the acceleration due to gravity (9.8 m/s²). This formula assumes that the strings are at a 90 degree angle to the shelf and there are no other external forces acting on the shelf.

3. What factors affect tension in strings holding up a shelf?

The tension in strings holding up a shelf is affected by several factors, including the weight of the shelf, the angle of the strings, the material and thickness of the strings, and the strength and stability of the wall or support where the strings are attached.

4. Can tension in strings be too high or too low?

Yes, tension in strings can be too high or too low. If the tension is too high, the strings may break or the shelf may be pulled too tightly against the wall, causing damage. If the tension is too low, the shelf may not be properly supported and could potentially fall.

5. How can tension in strings be adjusted?

Tension in strings can be adjusted by either changing the angle of the strings or by adjusting the length of the strings. Increasing the angle or decreasing the length will increase the tension, while decreasing the angle or increasing the length will decrease the tension. It is important to adjust the tension carefully to ensure that the shelf is properly supported and not at risk of falling.

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