Recent content by Lotus93

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?

I set them equal to zero because I thought the net force should be equal to 0 in order to hold the board still... is that wrong?

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]...