Block 1: T1 - Frictional Force = m*a
Block 3: m3*g - T3 = m*a
Pulley 2: (1/2)m2*r*a = (T3 - T1)*r
Then, T1 = m1*a + m1*g*u | T3 = m3*g - m3*a <--I plug these two into my 'Pulley 2'
I then algebraically solve for a and I end up with,
a = (m3*g - m1*g*u) / (0.5*m2 + m1 + m3)
torque has units of Nm. So I need some sort of distance I take it? Would I use the radius as this distance? Would it be Torque=(T3-T1)r ?
EDIT: This also doesn't seem to work. I'm feeling pretty lost at this point. =/
A block of mass m1 = 2 kg rests on a table with which it has a coefficient of friction µ = 0.53. A string attached to the block passes over a pulley to a block of mass m3 = 4 kg. The pulley is a uniform disk of mass m2 = 0.4 kg and radius 15 cm. As the mass m3 falls, the...
I also tried using the gravitational force as the Torque.
I used m*g*sin(20)=I*alpha
m*g*sin(20)/I=alpha and alpha*R = a
but this didn't work either.
I just solved this problem.
A certain non-uniform but cylindrically symmetric cylinder has mass 9 kg, radius 1.2 m, and moment of inertia about the center of mass 7.6 kg m2. It rolls without slipping down a rough 20° incline.
What is the acceleration of the cylinder's center-of-mass...
A 15 kg uniform disk of radius R = 0.25 m has a string wrapped around it, and a m = 5 kg weight is hanging on the string. The system of the weight and disk is released from rest.
a) When the 5 kg weight is moving with a speed of 1.7 m/s, what is the kinetic energy of the...
These are the hints they give me, I just don't understand at all what to do. I now believe though that the velocity I'm finding in the x-direction is incorrect, but I'm not sure why.
HELP: Because the cannon is moving, you must find the speed of the projectile with respect to the ground...
what do you mean "Then please label the 40 degrees"?
If you're implying that the answer is 40 degrees, I can assure with certainty that it isn't. I input these values into my homework online, and it returns with either an "OK" or "NO". I've tried 40 degrees, that was my first guess. =/
A circus cannon, which has a mass M = 5000 kg, is tilted at q = 40°. When it shoots a projectile at v0 = 80 m/s with respect to the cannon, the cannon recoils along a horizontal track at vcannon = 1 m/s with respect to the ground.
1. At what angle to the horizontal...
A bullet of mass 0.008 kg and initial speed 300 m/s penetrates an initially stationary pop can of mass 0.042 kg and emerges with a speed 210 m/s.
What is the initial momentum of the bullet and pop can system?
What is the final momentum of the bullet...
The two problems below are related to a cart of mass M = 500 kg going around a circular loop-the-loop of radius R = 15 m, as shown in the figures. All surfaces are frictionless. In order for the cart to negotiate the loop safely, the normal force exerted by the track on...
alright, thx. I was trying to do that last night, but gave up when I thought all i had was one side. I just used a previous triangle to find the angle, and then used that to find x. thanks for your help, I really appreciate your input. :)
A block of mass m = 3.5 kg rests on a frictionless floor. It is attached to a spring with a relaxed length L = 5 m. The spring has spring constant k = 12 N/m and is relaxed when hanging in the vertical position. The...
A spring is stretched a distance of Dx = 40 cm beyond its relaxed length. Attached to the end of the spring is an block of mass m = 8 kg, which rests on a horizontal frictionless surface. A force of magnitude 20 N is required to hold the block at this position. The force...
okay, so I'm assuming them I will be using the static coefficient because that's the kind tires use on pavement and whatnot. For the weight force... would that be the force pulling the car down the slope? m*g*sin(theta)?
or do u mean the weight force as 1000*9.81?
hello, yes, so i have my two frictional forces pointing upwards on the slope. I set it up like this... (1000)(9.81cos(theta))(.89)+(1000)(9.81cos(theta))(.61)-?=0 ...If this is correct, which I am not so sure b/c it just seems like I wouldn't be able to create an opposite force to equal it, so...
A car of mas m = 1000kg is sliding down a hill. The coefficients of friction between the car's tires and the ground are us=0.89 and uk=0.61. For what inclination angle will the car slide down the hill with a constant velocity?
The Attempt at a...
Hey, if someone could help me. I'm really stuck on this one, I just think I've dug myself a deeper hole trying to work this out. I would really appreciate any more help.
EDIT: I solved it. it's 10 N...