Mass hanging in a boxcar accelerating

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SUMMARY

The tension in the rope supporting a 1 kg mass in a boxcar accelerating at 7 m/s² is calculated using the equations of motion. The correct approach involves resolving forces into their components, leading to the conclusion that the tension is approximately 12 N. The user initially miscalculated the tension by incorrectly applying the Pythagorean theorem instead of using the correct force balance equations. The correct formula to find the tension is T² = (ma)² + (mg)².

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brake4country
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1. Homework Statement
A mass of 1 kg hangs inside a boxcar of a train. If the train accelerates at 7 m/s^2, what is the approximate tension in the rope?

A: 12 N

Homework Equations


F=ma

The Attempt at a Solution


I originally thought of doing this problem as pythagorean theorem, but I wanted to work out the x and y components for practice. Breaking up the components:
ΣFy=may
Ty-mg = may
Ty=10N

ΣFx=max
Tx=10*7
Tx=70N

Thus, T2=102+702
T=71 N

The problem sets it up just like I do but get 12 N for their answer:
T2=ma2+mg2

What am I doing wrong? Thanks in advance!
 
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brake4country said:
1. Homework Statement
A mass of 1 kg hangs inside a boxcar of a train. If the train accelerates at 7 m/s^2, what is the approximate tension in the rope?

A: 12 N

Homework Equations


F=ma

The Attempt at a Solution


I originally thought of doing this problem as pythagorean theorem, but I wanted to work out the x and y components for practice. Breaking up the components:
ΣFy=may
Ty-mg = may
Ty=10N

ΣFx=max
Tx=10*7
Tx=70N

Thus, T2=102+702
T=71 N

The problem sets it up just like I do but get 12 N for their answer:
T2=ma2+mg2

What am I doing wrong? Thanks in advance!
Can you show us your free body diagram for this problem? That is generally the first/best step on these problems. :smile:
 
brake4country said:
Breaking up the components:
ΣFy=may
Ty-mg = may
Ty=10N

ΣFx=max
Tx=10*7 ⇐
Tx=70N
Check the mass value used for the calculation of the x-component.
 

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