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Inclined plane with pulley problem help?
A 7 kg block is connected to a 13 kg block by a string and a pulley. The 7 Kg block is on the incline, while the 13 kg block is hanging. There is an angle of 38 degrees with the horizontal, and the co efficient of friction is 0.24.
Here's what I did:
For the hanging mass
Fnet = ma
Fg - Ft = 13a
127.5 - Ft = 13a
-Ft = 13a - 127.5
Ft = -13a + 127.5
For the 7 kg mass
Fnet = ma
-Fgx + Ft + Ff = 7a
- 42.2 + 127.5 +0.24 (7) (9.81) cos38 = 7a +13a
4.9 = acceleration
What am I doing wrong? It would amazing if you could help me!!!
1 Answer
- RockItLv 78 years agoFavorite Answer
mg = 7g
Mg = 13g
Mg - T = Ma, where a is the acceleration of the falling mass
T - mg*sin(38) = ma - Ff, where Ff = .24*mg*cos(38), the sign of Ff depends on whether the mass M is large enough.
eliminate T and solve for a.
Mg + Ff - mg*sin(38) = Ma + ma
a =( Mg + Ff - mgsin(38) )/(M+m)
substitute/calculate:
a = 3.61 m/s2
m moves up the incline, Ff points down the incline opposing this movement.
