How Does the Angle Affect Cable Tension in a Tow Truck Scenario?

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SUMMARY

The discussion focuses on calculating the cable tension in a tow truck scenario where a 1500 kg car is towed at a 23° angle to the horizontal while accelerating at 0.59 m/s². The tension in the cable can be determined using the formula F = (M * a) / cos(θ), where M is the mass of the car, a is the acceleration, and θ is the angle. The angle is crucial for decomposing the forces into horizontal (Fx) and vertical (Fy) components, allowing for accurate tension calculations.

PREREQUISITES
  • Understanding of Newton's Second Law of Motion
  • Basic knowledge of trigonometric functions, specifically cosine
  • Familiarity with force decomposition into components
  • Concept of acceleration in physics
NEXT STEPS
  • Study the application of Newton's Second Law in different scenarios
  • Learn about force decomposition and vector resolution techniques
  • Explore the effects of angles on tension in various towing scenarios
  • Investigate the role of friction in tension calculations
USEFUL FOR

This discussion is beneficial for physics students, engineers involved in mechanical design, and professionals in towing services seeking to understand the dynamics of cable tension in towing applications.

hofluff87
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A tow truck is connected to a 1500 kg car by a cable that makes a 23o angle to the horizontal. If the truck accelerates at 0.59 m/s2, what is the magnitude of the cable tension? Neglect friction and the mass of the cable.


i know that the car is accelerating at the same rate. my question is how do you include the angle into the equation for the cable tension?
 
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I think the purpose of the angle is so that you can de-compose the problems into components, Fx and Fy. Try to resolve both forces on the x-direction and y-direction first
 
I got it

F= (M * a)/cos (theta)
 

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