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This is the second of our many quizzes on Tension. Click on this to access more such competitive quizzes on Mechanics.

Tension Quiz 2

physics

A body is submerged and tethered to the bottom by a string. The body's density is smaller than that of water. The tension force generated in the string would then be,

Although the object appears to float, the string only allows it to do so for the length of the string. As a result, the buoyant force will exceed the body's weight, and the tension will equal the net force acting on it.

  The submerged body's density is smaller than that of water. As a consequence, it must float. It will not be able to float, however, because it is bound to the bottom by a string. Tension will develop inside the string as a result of the buoyant force pulling the body towards the surface. Since the net force acting is towards the top, the tension formed within the string will be against it. That is an indicator of a downward trend.<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/f2e7f8b4-46dd-4eaa-8d4f-965da1685cd74609951183577226146.png"/>

Tension developed inside the string attached to a free falling body before the string fully stretching will be equal to 

A free falling body may not be experiencing its weight throughout its motion but only until it fully stretches the string it is attached to. No tension will be developed inside any string or rope until it is fully stretched. 

  We must know that tension will be developed inside a stretched string. So, till before the body travelling a length of the string, it will not be stretched and so tension will not be developed.A free falling body won’t be experiencing its weight while falling but when it is attached to a string and it travels a length of the string so it is stretched completely, the motion of the body will have a sudden stop and a tension will be developed inside the string which is equal to the weight of the body.So, tension developed inside the string before it fully stretching will be zero. 

 A coolie is pulling a box placed on a platform using a rope. The weight of the box is 60 kg. The coefficient of friction between the box and the platform is 0.4. Then what will be the tension developed inside the rope just before the box starts moving? 

Tension developed inside the rope will be equal to the applied force and the applied force just before the box starts to move will be equal to limiting friction. So, tension will be equal to limiting friction. 

 $  \text{To just move the box, the coolie must overcome the friction first.}\\ \text{ So, just before motion, he will be applying a force equal to the frictional force between the ground and the box.} \\  \text{Therefore, an equal tension will be developed inside the rope also}\text{as the box is at the point of starting its movement but not actually moving.}\\  \text{So, the tension developed will be equal to the maximum frictional force.}\\ \Rightarrow T=F={{F}_{f}} \\  \Rightarrow T={{F}_{f}} \\  \text{We know, }{{F}_{f}}=\mu N \\  \Rightarrow T=\mu N \\  \Rightarrow T=\mu mg=\mu W \\  \therefore T=0.4\times 60=24N $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/f5a78234-16f8-472f-9b29-94c2b5d264981222453221606152196.png"/>

 A metallic string is being used by an elephant to drive a car. The car, which weighs a tonne, is on the handbrake, with a coefficient of friction of 0.7 between the tyres and the road. To prevent the elephant from dragging the vehicle, the driver applies a breaking force of 26 N. So, what is the minimum tension that the string should be able to withstand?

The minimum tension developed inside the string before the car starts to move will be equal to the sum of limiting friction and braking force. Limiting friction is given by the product of coefficient of friction and the normal force which will be equal to the car’s weight. 

 $  \text{Here, the string should have the capacity to hold both the limiting friction}\text{and the breaking force applied by the driver for successfully moving the car.}\\  \text{ Therefore, tension developed inside the string before motion will be,}\text{the sum of both limiting friction and braking force.}\\ \Rightarrow T={{F}_{f}}+{{F}_{b}} \\  \Rightarrow T=\mu N+{{F}_{b}} \\  \Rightarrow T=\mu W+{{F}_{b}}=0.7\times 1000+26 \\  \therefore T=700+26=726N \\  \text{Therefore the minimum tension the string must be capable of holding must be 726 N.}$  

A brick of mass $M$ is attached to a thick string having mass $m$ and length $L$. A force $F$ is pulling the brick away from the string. Then, what will be the tension at the point of contact of the brick and the string? 

 When using thick strings, the tension will not be equal at all times. The difference between the applied force and the tension on the point just before the point we took determines the tension at that point.Tension at a specific point on a thick string is given by the product of mass of the string till that point and acceleration of the string. I.e, $T=F-\left( \dfrac{mx}{L}\times a \right)$. 

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/c22de1be-dd2f-4ba0-b01d-a9d6bcccbe9e654594495669466551.png"/>$\\\text{ When we are using thick strings, the mass of the string is considerable. Because tension at all points will not be equal.}\\ \text{ Here, we are finding the tension at the point of contact between the brick and string.}\\  \text{ It will be given by the difference between the applied force}\text{and the sum of all tension forces excreted by the applied force on the points preceding it.}\\  \text{ Here, our point is at one end of the string.}\text{So, the tension excreted up to this point by the applied force on the string is given by,}\\  F'=m'\times a \\  a=\dfrac{\text{applied force}}{\text{total mass}}=\dfrac{F}{M+m} \\ \Rightarrow F'=\dfrac{mx}{L}\times \dfrac{F}{M+m} \\  \text{where“x”is the distance where we are considering the tension on the string from the end where external force is applied.}\\  \text{Here, x = L} \\  \Rightarrow F'=\dfrac{mL}{L}\times \dfrac{F}{M+m} \\  \therefore F'=\dfrac{mF}{M+m} \\  \text{Now, Tension at our desired point will be,} \\  \text{T = Applied force - Tension on the preceding point (}F') \\  \Rightarrow T=F-F'=F-\dfrac{mF}{M+m} \\  \Rightarrow T=F\left( 1-\dfrac{m}{M+m} \right) \\  \Rightarrow T=F\left( \dfrac{M+m-m}{M+m} \right) \\  \therefore T=F\left( \dfrac{M}{M+m} \right) $