- #1
undertow2005
i'm trying to learn this stuff on my own so please bear with me.
here's the problem:
you have two mass, a 10kg on a frictionless 35 degree inclined plane, which is attached to 20kg mass via a pulley cord to hang vertically. Whats the aceleration?
Solution:
for the box hanging freely they got this equation, 196N-T=20kg*acclerertion----i understand this
To get the equation for the box on the plane they did this:
98Nsin35=56N
98cos35=80N
out of these two results, 56n and 80N, they choose 56N to get the equation for the box which is T-56N=10kg*a
my question is why they they choose to use 56N and not 80N which would have resulted in the equation T-80N=10*a
continuing the solution the book has, they combined the two equations, 196N-T=20kg*a and T-56N=10kg*a to get this:
140N30kg*a
a=4.7 m/s/s I understand this part also.
here's the problem:
you have two mass, a 10kg on a frictionless 35 degree inclined plane, which is attached to 20kg mass via a pulley cord to hang vertically. Whats the aceleration?
Solution:
for the box hanging freely they got this equation, 196N-T=20kg*acclerertion----i understand this
To get the equation for the box on the plane they did this:
98Nsin35=56N
98cos35=80N
out of these two results, 56n and 80N, they choose 56N to get the equation for the box which is T-56N=10kg*a
my question is why they they choose to use 56N and not 80N which would have resulted in the equation T-80N=10*a
continuing the solution the book has, they combined the two equations, 196N-T=20kg*a and T-56N=10kg*a to get this:
140N30kg*a
a=4.7 m/s/s I understand this part also.