Newton's Second law: Tension on Cable

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aatari
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Hey guys is my solution correct for the question below?

1. Homework Statement

Consider a 5.0 kg watermelon that is being accelerated at 2.0 m/s2 [up] by a cable. Find the tension in the cable.

Homework Equations


Fcable = m.a

The Attempt at a Solution


Fcable = m.a
= 5.0 kg(2.0)
= 10N
 
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kuruman said:
Your solution is incorrect. Newton's second law says Fnet = ma. Fnet is the sum of all the forces acting on the watermelon, not just the force from the cable.
Ok I am confused now. I thought Fnet = m.a
 
Or in this situation should it be Fcable +Fgravity
 
kuruman said:
Yes, it should be as you say. Can you finish the problem now?
Fcable = 10 N
Fgravity = -49N

Fcable + Fgravity = 10N - 49N
= 39 N [down]

Is this correst?
 
How do you figure Fcable = 10 N? It seems that you multiplied the acceleration by the mass and called that Fcable. We agreed that mass times acceleration is the net force. Write an expression for the net force as the sum of all the forces, set it equal to mass times acceleration and solve for Fcable.
 
kuruman said:
How do you figure Fcable = 10 N? It seems that you multiplied the acceleration by the mass and called that Fcable. We agreed that mass times acceleration is the net force. Write an expression for the net force as the sum of all the forces, set it equal to mass times acceleration and solve for Fcable.
I think I get it now.

So Fnet = Fcable + Fgravity
10N = Fcable - 49N
10N + 49N = Fcable
Fcable = 59N
 
kuruman said:
Yes, you got it. :smile:
Thanks for your help.