Incline angle for accelerating cart and Tensions in strings for a suspended mass

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SUMMARY

The discussion focuses on solving two physics problems involving a trolley on an incline and a suspended mass. For the first problem, the incline angle is calculated to be 23.58° using the formula Sinα = a/-g, where a is the acceleration of the trolley at -4 m/s² and g is the gravitational acceleration at 10 m/s². In the second problem, the tension in the strings supporting a 12 kg weight is determined, resulting in T2 and T3 both equaling 69.3 N, derived from the equilibrium equations involving the weight's tension and angles.

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  • Understanding of Newton's laws of motion
  • Familiarity with trigonometric functions and their applications in physics
  • Knowledge of tension in strings and equilibrium conditions
  • Basic grasp of kinematics, specifically acceleration and velocity
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  • Study the derivation of incline angles in physics problems
  • Learn about tension in multi-string systems and their equilibrium
  • Explore advanced kinematics involving frictionless surfaces
  • Investigate the effects of different angles on tension and acceleration
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of incline problems and tension calculations in static systems.

JAZZ541
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Homework Statement


Hi all, I have solved these problems just wanted to see if it's all ok

problem 1:
a trolley starts at the lowest point of a diagonal plane with a 20m/s speed. Its acceleration is 4m/s^2 going downhill, the trolley goes up then slows until it reaches 0 speed and then goes down the plane with the same acceleration
a. what is the slopes incline? Its is known that the trolley moves without friction?

problem 2:
A weight that has m = 12Kg is pendent in rest on a string attached to two other strings and creating an identical β = 60° with the ceiling, calculate the tension of all strings

Homework Equations

The Attempt at a Solution



solution 1:
[/B]
upload_2016-9-18_20-38-33.png
V0 = 20m/s

a = -4m/s2

g = 10m/s2

m1g • sinα = m1a

N = m1g • cosα

-m1g • sinα = m1a

Sinα = a/-g

Sinα = -4m/s2/-10m/s2

α = 23.58°solution 2:

upload_2016-9-18_20-39-23.png
T1 = mg

m = 12Kg

mg = T1 = 120N

T1 = T2cos30° + T3cos30°

T1 = cos30° • (T2 + T3)

120N = cos30° • (T2 + T3)

T2 + T3 = 120/cos30° = 138.56

T2sin30° = T3sin30°

T2 + T3 = 138.56

2T = 138.56

T2 = T3 = 69.3

TY!
 
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Both solutions look OK.
 
Ty very much!
 

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