KA force of tension question

[howtomedicine]

The answer was 147N but I'm not sure why. If the person is suspended by 4 ropes at an angle, shouldn't the tension in a single rope be (1/4)mg • sin[angle]

The hints stated that we were given the vertical component already but I'm not understanding how.


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[aldol16]

The answer was 147N but I'm not sure why. If the person is suspended by 4 ropes at an angle, shouldn't the tension in a single rope be (1/4)mg • sin[angle]

Draw a free body diagram. Those are always very helpful for problems like these. Okay, so it tells you that there are four ropes and all four ropes have the same angle. Therefore, they each bear 1/4 of the total weight. That's one critical piece of the puzzle. That means that each rope only "feels" 60 kg/4, or 15 kg each. The second key piece is that this problem is not asking for the tension in the rope but rather the vertical component of force acting on a single wire. In your free body diagram, you should only have one arrow pointing down - that of the weight of the object. So you already know the vertical component of force. That is, the vertical component of force in this entire system points down with a magnitude of 60 kg*9.8 m/s^2. Now remember what I said earlier. Since there are four wires, each wire bears 1/4 of the total force. Thus, each wire has a vertical component of force of 60*9.8/4.

 

[howtomedicine]

Ahhhh I kept reading it and interpreting as them asking for the vertical component of tension, thanks a lot!!


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[aldol16]

Ahhhh I kept reading it and interpreting as them asking for the vertical component of tension, thanks a lot!!

It is the vertical component of tension!

 

[howtomedicine]

So if we were given the tension of a single rope, then the vertical component would be Tsin(angle) which would be equal to 1/4mg?


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[aldol16]

So if we were given the tension of a single rope, then the vertical component would be Tsin(angle) which would be equal to 1/4mg?

If you were given the tension force in a single rope, then yes, the vertical component would be T*sin(theta) where theta is the angle the rope makes with the horizontal. And yes, that would equal 1/4*m*g in this case. In fact, you can easily calculate the tension force in this case in each rope. As I said, the vertical component is 60*9.8/4, which is about 150. Therefore, the tension must be 150/sin(theta).

 

[howtomedicine]

Thanks a lot for the further clear up!


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[LoveBeingHuman:)]

Sometimes I feel like half the battle of the MCAT is not making careless mistakes