Recent content by KKazaniecki

torque is : r*F*sin(theta).where both the tensions are 90 degrees from the Axis. so sin(90)=1. here the torque is only r*F. where r is the radius. And I'm having trouble working out what the force is. I know it's somehow connected to the two tensions. Where the tension due to block 1( with mass...

the , I think I need few more hints. well I have no idea how to relate the two tensions to the torque. So I tried using the equation of : a(linear)=r*a(angular). So since the rope doesn't stretch, meaning that if the m goes down by say 1cm the 3m block will go 1cm to the right. So this means...

would it be, m*g-m*a=t1. t1-T(torque)=3*m*a? rearranging we get T(torque)=m*(g-4a). And I could keep working on it to get a. then just do T(torque )/ I = a(angular) . but is the relation between the tensions and torque correct there ?

mhmm, so how would I go about solving it with two different tensions?

1. Homework Statement The system shown in the diagram contains two blocks, of masses 1.9 kg and 5.7 kg, connected by a light string over a pulley of radius 0.15 m and rotational inertia 2.8 kg m 2 . The block of mass 5.7 kg is free to slide on a horizontal frictionless surface and the pulley is...

ok so , mgh + mgx = 1/2kx^2 . In other words Gravitational Potential = Spring Potential. Which as you said is the one where it's above the spring , and when it's compressing the spring. So Energy is conserved . Rearranging that to find x , We get: x = (sqrt(g m (g m+2 h k))+g m)/k. Where k =...

what I meant is that . It's one or the other. So the energy potential at the maximum height is the same as the energy potential of the spring when it is compressed. Because Energy is conserved . It goes from Gravitational Potential --> to kinetic ( as it falls ) ---> to spring potential ( when...

1. Homework Statement A spring-like trampoline dips down 0.07 m when a particular person stands on it. If this person jumps up to a height of 0.35 m above the top of the uncompressed trampoline, how far will the trampoline compress after the person lands? 2. Homework Equations...

hi all, here's the problem "A dog sits on the left end of a boat of length L that is initially adjacent to a dock to its right. The dog then runs toward the dock, but stops at the end of the boat. If the boat is H times heavier than the dog, how close does the dog get to the dock? Ignore...

thanks guys for your help, I understand it now :)

thanks , that works .Could you explain why ? isn't acceleration due to gravity downwards , and the system acceleration upwards . So why do they add up together?

hey, I'm currently going through a mechanics course on edx.org . And part of last's week homework was this problem. well okay I don't know how to add images so I'm going to try my best to describe it. There is a platform has a mass of 200kg. From that platform a rope hangs( with tension TA...