Homework Statement
My last inclined plane problem has a second part to it: There are some energy problems I need to solve.
A double pulley consists of two separate pieces that are welded together; a hoop of radius 10.0 cm and a disk of radius 15.0 cm. The moment of inertia of the pulley is...
Based on what you all have said, I figured I could get a Newton value by multiplying the breaking strength by the cross-sectional area of the wire. Cross-sectional area would be m2 so the units would cancel out.
F_4=110.0*106*(\frac{0.0082}{2})^2*\pi*\frac{\sqrt{39}}{8}=45347.42...
Homework Statement
I want to convert 110.0 x 10^6 N/m^2 to Newtons. How do I do this?
Reason: I have the following problem and the following question to answer:
P: A heavy mass is supported by a beam-cable system as shown in the figure. The beam is 5.000 m long with a mass of 36.00 kg. The...
Alright, got an email from my professor. Apparently there was an error in the problem. The radius of the larger pulley is now 15cm instead of 25cm.
α=\frac{9.81*(0.1*24-0.15*10*(\sin35°+0.25\cos35°))}{0.16+10*0.15^2+24*0.1^2}=19.3446
And using 0.15 instead of 0.25...
Apparently I got the wrong number the first time. It's not α = 3.33055. It's α = 4.34596.
α=\frac{9.81*(0.1*24-0.25*10*(\sin35°+0.25\cos35°))}{0.16+10*0.25^2+24*0.1^2}=4.34596
This is so confusing. It seems like such a simple question: "Which direction are they going?" I never would have...
A minor thing I realized: I was using centimeters rather than meters in the last equation. So the latest answer for a1 is a1 = -0.95.
Sorry, I just realized this, but you said in your 2nd-to-last post that the blocks cannot start moving from rest. However the problem states "At the instant...
It's more likely that I came up with the incorrect equation than that my professor gave a bad problem. I copied the problem exactly as-is.
Maybe if I try solving it your way I will get a positive answer. (Which is apparently what I am supposed to get?) What am I doing wrong exactly 1-post-back...
I tried doing this but I couldn't eliminate the tensions. In fact it ended up looking like a lot more work than the last equation I posted, so I guess I'm not doing it right:
R * m1 * a1 = (R * T1) - (R * μ * m1 * g * cos(θ)) - (R * m1 * g * sin(θ))
r * m2 * a2 = (r * m2 * g) - (r * T2)
R * T1...
Here's how I did it:
Equations of motion:
m1 * a1 = T1 - (μ * m1 * g * cos(θ)) - (m1 * g * sin(θ))
m2 * a2 = m2 * g - T2
I * α = T2 * r - T1 * R
Deduced from equations:
α = a1 / R = a2 / r
Sub a2 with (a1 * r / R):
m2 * a1 * r / R = m2 * g - T2
Sub α with (a1 / R):
I * a1 / R = T2 * r - T1 *...
Okay. So this will become relevant when I calculate the blocks' accelerations.
Also, I am now factoring in the moment of inertia and the radii. For the acceleration of m1 I am now getting:
a1 = ( (m2 * g * r / R) - (μ * m1 * g * cos(θ)) - (m1 * g * sin(θ)) ) / ( m1 + (m2 * (I/R)^2) +...
Homework Statement
A double pulley consists of two separate pieces that are welded together; a hoop of radius 10.0 cm and a disk of radius 25.0 cm. The moment of inertia of the pulley is 0.160 kg-m^2. A 15.0 kg block (m1), on a 35.0° incline, is attached to the outer pulley by a massless...