- #1
byronsakic
- 17
- 0
Hello, here is the question.
Two masses, 4.0 kg and 6.0 kg, are connected by a “massless” rope over a “frictionless” pulley as pictured in the diagram. The ramp is inclined at 30.0º and the coefficient of kinetic friction on the ramp is 0.18.
http://www.clickandlearn.org/Physics/SPH4U/tests/ch2_test_files/i0430000.jpg
(b) Determine the acceleration of the system once it begins to slide.
i only need help on b.. finding the acceleration. if i can get the right acceleration i can get all the other parts.
Now i solved it by doing this. making 2 equations.
4 kg = m1 6kg = m2 motion to the right is positive. uk = 0.18
m1a = T - [Ff + Fg||]
m1a = T - (uk)(m1g)cos30 - sin30m1g
m1a = T - 6.1 - 19.6 equation 1
m2a = m2g - T
m2a = 58.8 - T equation 2
add the two equations together you get. (T cancel out)
a (m1 + m2) = 58.8- 25.7
a = 33.1 / 10
a = 3.3 m/s ^2
i have been lookin over my calculation over and over and the acceleration is different from the answers.
here is the SOLUTION:
(b)
For the 4.0-kg mass:
http://www.clickandlearn.org/Physics/SPH4U/tests/ch2_test_files/a0430000.jpg
4.0 kg(a) = T – uKmg(cosX) – mg(sin X)
4.0 kg(a) = T – 13.5 N
For the 6.0-kg mass:
http://www.clickandlearn.org/Physics/SPH4U/tests/ch2_test_files/a0430001.jpg
6.0 kg(a) = 58.8 N – T
Solving the system of equations:
a = 4.5 m/s2
according to the solution... for equation 1.. it is + 6.1 - 19.6 = -13.5
but it doesn't make sense because it should be -6.1 - 19.6 = -25.7
is there a trick I am missing or do you think the solution is wrong? this question is vital for a test on monday.
thank you
byron
Two masses, 4.0 kg and 6.0 kg, are connected by a “massless” rope over a “frictionless” pulley as pictured in the diagram. The ramp is inclined at 30.0º and the coefficient of kinetic friction on the ramp is 0.18.
http://www.clickandlearn.org/Physics/SPH4U/tests/ch2_test_files/i0430000.jpg
(b) Determine the acceleration of the system once it begins to slide.
i only need help on b.. finding the acceleration. if i can get the right acceleration i can get all the other parts.
Now i solved it by doing this. making 2 equations.
4 kg = m1 6kg = m2 motion to the right is positive. uk = 0.18
m1a = T - [Ff + Fg||]
m1a = T - (uk)(m1g)cos30 - sin30m1g
m1a = T - 6.1 - 19.6 equation 1
m2a = m2g - T
m2a = 58.8 - T equation 2
add the two equations together you get. (T cancel out)
a (m1 + m2) = 58.8- 25.7
a = 33.1 / 10
a = 3.3 m/s ^2
i have been lookin over my calculation over and over and the acceleration is different from the answers.
here is the SOLUTION:
(b)
For the 4.0-kg mass:
http://www.clickandlearn.org/Physics/SPH4U/tests/ch2_test_files/a0430000.jpg
4.0 kg(a) = T – uKmg(cosX) – mg(sin X)
4.0 kg(a) = T – 13.5 N
For the 6.0-kg mass:
http://www.clickandlearn.org/Physics/SPH4U/tests/ch2_test_files/a0430001.jpg
6.0 kg(a) = 58.8 N – T
Solving the system of equations:
a = 4.5 m/s2
according to the solution... for equation 1.. it is + 6.1 - 19.6 = -13.5
but it doesn't make sense because it should be -6.1 - 19.6 = -25.7
is there a trick I am missing or do you think the solution is wrong? this question is vital for a test on monday.
thank you
byron