find the tention in block at distance x form the end where f is applied
Answers
T = F * (L-X)/L
I’m going to elaborate but it is not necessary because the distribution of F along the rope must be a linear function whose value is F at one end and zero at the other.
For those who do not think so, let me elaborate. The rope placed horizontally because it never mentions gravity ( and if someone regards gravity why not an angle alpha and why vertical?).
A rope definitely has got a mass M. If you pull from an end with a force F,
F = M a => a = F/M
if we consider the rope at a middle point distant x from the end of F, the mass remaining behind is (L - x )* M/L
This mass is being dragged by the rest of the rope by virtue of the tension T(x) and is being dragged at the same acceleration as the rope (assuming it of fixed length). So the second law for this remaining part of the rope is T = m a
T(x) = (L-x) M/L * F/M
the M cancel out and T(x) is found.
T is constant when it is constant but not in this situation.
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If there is a T tension in both sides of a rope, what is the tension in the rope?
A string of length L and mass M is lying on a horizontal table. A force F is applied at one of its ends. What is the tension in the string at a distance y from the end at which the force is applied?
How do we find tension if one end of the rope is pulled by 120N force and other end by 100N?
You have a string of length L. You pull the string from end A with force F. The other end is free. What is the tension "l" distance away from end A?
What will be the tension in the rope that is pulled from its ends by two opposite forces 100N each?
At a distance x from its end we took a small element dx
Mass of dx= Mdx/l =dm
Rope is moving with an acceleration a
Equation of motion for the small element dx:
{T + dT} - T = dm × a
dT = [M dx/l] × F/M
dT = F dx /l
Integrating both the sides with limit of T being F to F(x) and limits of x being x to 0 we get,
F(x) -F = - Fx/l
F(x) = F[1-x/l]
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Let the magnitude of force exerted by the portion of the rope to the left of position x be F(x), and F(x+dx) be the magnitude of the force exerted by the rope to the right of position (x+dx). Then, the net force acting on element dx of the rope is F(x)−F(x+dx), which by Newton’s second law of motion is equal to MLdxFM, i.e., F(x)−F(x+dx)=MLdxFM, or F(x+dx)−F(x)dx=−MLFM=−FL. So, we have dFdx=−FL, or F(x)=−FLx+constant. The constant is evaluated by using the fact that at x=0, F(x)=F, the applied force. So, the force acting on the rope at a distance x from the end where the force F is applied is F(x)=−FLx+F=F(1−xL).
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Assume the rope to be having a linear mass distribution. Divide it into several tiny particles now.
Now all what you have to do is to observe in your mind what’s happening in this system.
Particle at one end has to carry whole of the rope and is being pulled by force F. Particle at the other end has to carry none. Particle in middle has to carry the next half. Gut feeling will tell you that the function, force here, is linear in nature. So you can now construct the equation all by yourself and arrive directly at the final result—-
F(x) = F*(x/L) where x is measured from the other extreme end.
Measuring from the front end it would be F*(L-x)/L
:)
Its good to use mathematics in physics as it is the language of science. But you must also be able to use your understanding well and mathematics as less as possible while arriving at the same result.