This is what I have now...
(-0.5)(m)(g)(cos(30))+(B)(0.4)=0
B=-(-.5)(45)(9.8)(cos(30))/0.4
B= 477.4
Fx-B(sin(30))=0
Fx=(477.4)(sin(30))
Fx= 238.7
Fy-mg+B(Cos(30))=0
Fy=-477.4(Cos(30))+(45)(9.8)
Fy=27.6Am I anywhere close to being right?
This is the problem:
http://oi47.tinypic.com/6qij4o.jpg
I set fx, fy, and B equal to zero.
'x' forces: Fx - Bsin(30) = 0
'y' forces: Fy - mg + B(cos(30)) = 0
lower end: -0.5*mg(cos(30)) + 0.4*B = 0
Using the third equation I solved for B, and got -477.4.
I then plugged B into the...
I've listed the whole problem because I know I'll have more questions, but all I want to know atm is whether I'm plugging the right values into the equation. Thanks in advance for your help.
Homework Statement
A compact disc speeds up uniformly from rest to 5200rpm in 620rad. The disc’s...
I spoke with one of my classmates, and this is what we worked out...
v^2 = vo^2 + 2ay
3^2 = 2(a)(0.5)
a = 9m/s
Net Torque = I * alpha
a = alpha * r
Trsin90 = I (a/r)
T(0,072)= I (9/0.072)
F = ma
T - mg = -ma
T = mg - ma
T = 1.3(9.8) - 1.3(9)
T = 1.04
(1.04)(0,072)= I...
Homework Statement
A 1.3 kg block is tied to a string that is wrapped around the rim of a pulley of radius 7.2cm. The mass of the pulley is 0.31kg. The block is released from rest. If the velocity of the block is 3.0m/s after it falls 0.5m, what is the moment of inertia for the pulley?[/b]...