Static Equilibrium of a 10m Rod with a Hanging Weight at 1.46m Distance

In summary, the conversation discusses the mechanics of a uniform rod being supported by a cable and held against a wall by friction. The task is to determine the minimum distance from the left end of the rod at which a block can be hung without causing the rod to slip, given certain weight and angle conditions. The solution involves using equations to find the necessary tension and friction forces, and ultimately calculates a minimum distance of 1.46 meters.
  • #1
nameVoid
241
0
one end of a uniform 10 meter long rod weighing 50 Newtons is supported by a cable at an angel of 53 deg with the rod the other end rests against the wall where is is held by a friction us=2 . determine the minimum distance x from the left end of the rod at which a block weighing 20 Newtons can be hung without causing the rod to slip.

IS THIS CORRECT?

FX=n-Tcos53=0
FY=Tsin53+FS-70=0
FT=10Tsin53-250-20x=0
20x=10Tsin53-250
2x=Tsin53-25
FS=2Tcos53
Tsin53+2Tcos53=70
T=70/ sin53+2cos53
2x=70sin53/ sin53+2cos53 -25
x=35sin53/ sin53+2cos53 -25/2=1.46m
 
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  • #2
nameVoid said:
one end of a uniform 10 meter long rod weighing 50 Newtons is supported by a cable at an angel of 53 deg with the rod the other end rests against the wall where is is held by a friction us=2 . determine the minimum distance x from the left end of the rod at which a block weighing 20 Newtons can be hung without causing the rod to slip.

IS THIS CORRECT?

FX=n-Tcos53=0
FY=Tsin53+FS-70=0
FT=10Tsin53-250-20x=0
20x=10Tsin53-250
2x=Tsin53-25
FS=2Tcos53
Tsin53+2Tcos53=70
T=70/ sin53+2cos53
2x=70sin53/ sin53+2cos53 -25
x=35sin53/ sin53+2cos53 -25/2=1.46m
It looks good to me. :approve:
 

1. What is meant by "static equilibrium" in this scenario?

In this context, static equilibrium refers to the state in which the 10m rod with a hanging weight at 1.46m distance is not moving or rotating. This means that the forces acting on the rod are balanced, resulting in a stable and stationary position.

2. How is the distance of 1.46m determined?

The distance of 1.46m is calculated as the distance from the pivot point (where the rod is attached) to the center of mass of the hanging weight. This distance is important in determining the torque, or rotational force, acting on the rod.

3. Why is the weight of the hanging weight relevant in this scenario?

The weight of the hanging weight is relevant because it is one of the forces acting on the rod. In order for the rod to be in static equilibrium, the weight of the hanging weight must be balanced by an equal and opposite force, such as the tension in the rod.

4. How does the length of the rod affect the static equilibrium?

The length of the rod affects the static equilibrium by determining the distance between the pivot point and the hanging weight. A longer rod will result in a larger distance, which can change the torque and therefore the forces needed to maintain static equilibrium.

5. What other factors can affect the static equilibrium of the rod and hanging weight?

Other factors that can affect the static equilibrium include the material and thickness of the rod, the shape and weight distribution of the hanging weight, and the strength and direction of external forces such as wind or vibrations.

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