Recent content by shake94

Okay sure. Please let me know. I'll be waiting for your reply. I'll try my best to understand it myself as well.

The reason why I am asking is because of this explanation my professor gave me. He showed the connection between the accelerations of the two blocks like this and I didn't understand how. This is the actual figure to the explanation above. It's different from this problem I posted here. He...

So the acc of 20 KG BLOCK= - acc of 10 kg block? Is that right? And is the connection between the tensions right?

Still doesn't ring a bell. I mean, if 10kg block is moving down by x then shouldn' that be -ve. And 20 kg is move upwards by x and that has to be positive. Either way the equation looks fine. Until the part where you brought up the derivative. Acceleration is change in velocity over time and...

The only way I see the motion is that if the 10 kg block moves, then the 20 kg block moves as well. So let's say the 10 kg block moves a distance x meters, then the pulley at the top, pulls the pulley on the left side that moves the 20 kg block upwards. This is how I see it. So if the 10 kg...

Please explain a little more further. It's still not clear to me. If I move the block with m=10 kg a distance of x meters DOWNWARD, then the 20 kg will move x meters upwards as well, is that what you're trying to say here? So acceleration of 10 kg= twice the acceleration of 20 kg? I mean how...

Homework Statement Look at figure 4 Two blocks are attached to mass-less ropes and pulleys that have negligible mass and negligible friction. a. Find the connection between the tensions in the ropes. b. Find the connection between the accelerations of the blocks. c. Calculate...

Okay. I got it! Thank you so much! I really appreciate it! :smile::smile:

He can change the direction, or accelerate. ?

I understand your poitn about the trig functions, I have understood them because they're a slope so I will have to change the angles as well. Thank you for confirming that. Okay, so the new θ I find from substituting the equations that way will be the θ that is required to "break loose" the...

Well, to show that the crate is not about to slip, I need to show that the component of gravity hasn't yet reached Fsmax. And that's my attempt here. And it breaks loose just when the component of gravity can no longer oppose it. So, how do I find θ value for when it breaks loose?

A flatbed truck and Static Friction and Crate "breaking loose" Homework Statement A flatbed truck is going downhill on a slope of 10 degrees at a constant speed of 20 m/s (a little over 50mph). A crate of 500 kg is sitting on the flatbed , without being attached by ropes or cables. The...

Your procedure seems right. Hopefully you included all other forces such as the Normal and the Friction into your calculation.