A gymnast with mass 50 kg suspends herself from the lower end of a hanging rope of negligible mass. The upper end of the rope is attached to the gymnasium ceiling. (a) What is the gymnast's weight? (b) What force (magnitude and direction) does the rope exert on her? (c) What is the tension at the top of the rope?
Answers
Answer:
A gymnast with mass 50 kg suspends herself from lower end of a rope. the upper end of rope is attached to the ceiling. what is the tension at the top of the rope?
(a)The weight of the gymnast is 490N, (b) The force, which the rope exerts on the gymnast is 490N, upwards, and (c) The tension at the top of the given rope is 490N.
Given,
The mass of the gymnast is 50 kg, she suspends herself from the lower end of a hanging rope of negligible mass.
The rope's upper end is attached to the gymnasium ceiling.
To Find,
The weight of the gymnast.
The force, which the rope exerts on the gymnast.
The tension at the top of the given rope.
Solution,
We can solve this mathematical problem using the following method.
The product of the gymnast's mass and the acceleration due to gravity determines the magnitude of her weight.
∴w= m×g. [Where g =9.8m/]
∴w=50×9.8=490N.
The gymnast's gravitational force which is actually her weight is in the negative y-direction.
A force is applied upward.
Her weight has already been calculated.
As a result, the rope pulls the gymnast up with a force of 490 N.
As given, the rope's mass is negligible, therefore we assume this is weightless.
Now on the rope, only two forces will work - the upward force by the celling, and the downward force by the gymnast.
The net vertical force on the rope in equilibrium must be zero, according to Newton's first law.
Also, the tension at the top of the rope is equal to the weight on the rope.
∴T=W=490N.
Hence, (a)The weight of the gymnast is 490N, (b) The force, which the rope exerts on the gymnast is 490N, upwards, and (c) The tension at the top of the given rope is 490N.
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