The systems shown below are in equilibrium (with m = 5.50 kg and θ = 28.0°). If

Physics
Tutor: None Selected Time limit: 1 Day

The systems shown below are in equilibrium (with m = 5.50 kg and θ = 28.0°). If the spring scales are calibrated in newtons, what do they read? Ignore the masses of the pulleys and strings and assume the pulleys and the incline are frictionless.


May 31st, 2015

By looking at and analyzing the position of the masses in the diagram, the universal g constant, 9.81 m/s^2, must be divided by 2 and multiplied by the total kg:
(9.81 m/s^2) / 2 = 4.905 m/s^2
4.905 m/s^2(11.00 kg) = 49.1 kg(m/s^2) = 49.1 N 


Since there are 2 masses:
2(5.5 kg) = 11.00 kg

Now to get the answer, multiply the total kg by the universal g constant, 9.81 m/s^2 because the force is acting downwards:
11.00 kg(9.81 m/s^2) = 98.1 kg(m/s^2) = 98.1 N

May 31st, 2015

Did you know? You can earn $20 for every friend you invite to Studypool!
Click here to
Refer a Friend
...
May 31st, 2015
...
May 31st, 2015
Dec 10th, 2016
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer