When the force at the ends of a bar pushes rather than pulls, the bar experiences?
1 answer:
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Answer:
A) 1 N
B) 50 N
Explanation:
Let us consider that the string does not deform.
To solve this problem lets consider the whole system as two parts. In the initial case, the first part will be de 100N being exterted to the whole system and in the second the 100 blocks system.
In this case we can imagine as the whole system being pulled by 100 N, and therefore its acceleration will be:
a = 100 N /(100 m)
where m stantds for the mass of one block
Now, the whole system and its individual parts must move with the same acceleration otherwise the string would stretch.
Now lets consider the first part of the system as the first block, and the second part as the other 99 blocks.
The Tension of the string pulling the 99 blocks must be so that it exterts the enough force to move that 99blocks-system at an acceleration a, since that sub-system has a mass of 99m
T1 = 99 m * a = (99 m) * (100 N/ 100 m) = 99 N
Now lets consider an intermidiate sub-system, where the first part is made of n blocks and the second susbsystem is made of (100 -n) blocks
Following the same logic, the tension of the corresponding string must be the acceleration of the whole systems times the mass of the second subsystem:
Tn = (100 -n)m * ( 100 N / 100 m ) = (100 -n) N
a)
Therefore the tension in the string connecting block 100 to block 99 must be
<u>T99 = 1 N</u>
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b)
And
<u>T50 = 50 N</u>
