How Do You Calculate Tension in a Lifting System with Angles and Acceleration?

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SUMMARY

The discussion focuses on calculating tension in a lifting system involving a 3.5 Mg engine suspended by a spreader beam and lifted by a crane with an acceleration of 4 m/s² and a velocity of 2 m/s. The user initially attempted to apply the formula F=ma but encountered errors in their calculations. The correct approach involves using a Free Body Diagram and the equation ΣF_y = m * a_net to determine the tensions in the chains CA and CB, which are both found to be 27.9 kN.

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Knowledge of Free Body Diagrams
  • Familiarity with trigonometric functions (sine and cosine)
  • Basic concepts of dynamics in engineering
NEXT STEPS
  • Study the principles of Free Body Diagrams in engineering mechanics
  • Learn about tension calculations in static and dynamic systems
  • Explore the effects of angles on force components in lifting systems
  • Review advanced dynamics topics, including acceleration and velocity relationships
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Engineering students, particularly those studying dynamics, mechanical engineers, and professionals involved in lifting system design and analysis.

erica715
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This is actually for an engineering dynamics class but its basically physics. I know how to find force using F=ma. But I can't figure out how to find tension given velocity and acceleration. My book is no help :(

1. A 3.5 Mg engine is suspended from a spreader beam AB having a negligible mas and is hoisted by a crane which gives it an acceleration of 4 m/s^2 when it has a velocity of 2m/s. Determine the force in chains CA and CB during the lift. To help you visualize AB represents the distance between where the two ropes connect to the top of the engine and C is where the two ropes meet together at the crane. The angle between line AC and AB is shown as 60 degrees.


Homework Equations


F=ma


The Attempt at a Solution


First I tried the obvious F=ma. 3500kg(9.81)(4)=137340. Wrong.

Then I tried...T-cos(60)(3500)(9.81)-sin(60)(3500)(9.81)=(3500)(4)
T=26567.48. Wrong.

I feel like I need to include the velocity that's given but I have no idea where.

The answer in the back of the book says that T(of CA) and T(of CB) are both 27.9 kN.

Ugh I've been at this for a couple hours now and I only did 2 of my homework problems so far. Help please!
 
Physics news on Phys.org
There are two wires which have a tension T. Draw a Free Body Diagram of the problem and list the forces.

Using \Sigma\vec{F}_y=m\vec{a}_{net}, solve for T.
 

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