How much tension must a rope withstand if it is used to accelerate a 1400 kg car

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SUMMARY

The tension required for a rope to accelerate a 1400 kg car vertically upward at 0.50 m/s² is calculated using Newton's second law, F=ma. The gravitational force acting on the car is 13720 N (calculated as mg, where g is 9.81 m/s²). The total upward force, which is the tension in the rope, must counteract both the weight of the car and provide the necessary acceleration. Therefore, the tension in the rope is the sum of the gravitational force and the force due to acceleration, resulting in a final tension value of 14420 N.

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Knowledge of gravitational force calculation (Fg=mg)
  • Basic concepts of forces acting in different directions
  • Familiarity with units of force (Newtons)
NEXT STEPS
  • Study the application of Newton's laws in various scenarios
  • Learn about tension forces in different contexts, such as pulleys and cranes
  • Explore advanced dynamics involving multiple forces and accelerations
  • Investigate real-world applications of tension in engineering and physics
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Students studying physics, particularly those focusing on mechanics, engineers involved in design and analysis of lifting systems, and anyone interested in understanding forces in motion.

mortho
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[SOLVED] Rope Forces

Homework Statement


How much tension must a rope withstand if it is used to accelerate a 1400 kg car vertically upward at 0.50 m/s2?



Homework Equations



F=ma

The Attempt at a Solution



Well i know that the force going up is going to be 700 N and i need to somehow subtract it from a force going horizontally to get my answer. I just don't know which..maybe I'm just missing something that i can 't think of ..NEED HELp thanks!
 
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What are the forces acting vertically? Let's say we pick upward to be positive, what's the equation for Newton's Second Law in terms of the summed forces?
 
Fg and Fn are acting vertically. the equation for Fg is Fg=mg and so i got 13720. Fg must be equal to Fn because of the flat ground so would that be that answer. But then the acceleration's going to affect it so...then 700 N is going to play a part right? Do i subtract it?? I'm so confused!
 
Ok, so you're going to have the force on the car due to gravity, this is the car's weight. But I believe you're mistaken about the second force. We're not going to have a normal force, as the car won't be on the ground once it accelerates vertically. What force is pulling the car upwards, where does it fit into our net force equation?

You've already identified the weight force, now we need one more:

\Sigma F_{vert} = (Upward force) - mg = ma
 
i don't understand the upward force part..would we do this 700-13720 because 700 is what was pushing down but since it's lifted it lessens?
 
The upward force is the tension in the rope. Imagine the car hanging from a crane (or something). If it is being pulled upwards, there are two forces present. The tension, and gravity. They are acting in different directions. You need to add them up according to Newton's second law to get the net force, which will equal ma. You know the mass and acceleration, all you need to do is solve for the tension (or the upward force as hotcommodity called it). Look at post #4 carefully.
 
so i don't subtract 700 from 13720? i would add them instead?
 
mortho said:
so i don't subtract 700 from 13720? i would add them instead?

Yes, that's right.
 
and that would be my final answer? wow that's huge!
 
  • #10
Yes, that's what your answer will be.
 
  • #11
THANK YOU ! oh by the way i sent you a message in your inbox thing!
 
  • #12
Got it.
 

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