Motion and Acceleration of Connected Blocks on an Inclined Plane with a Pulley

  • Thread starter Thread starter moondawg
  • Start date Start date
  • Tags Tags
    Blocks Pulley
AI Thread Summary
The system of two connected blocks on inclined planes will move to the left, driven by the heavier 100 kg block. The calculated forces show that the 100 kg block exerts 490 N down the ramp, while the 50 kg block exerts 391.8 N down its slope. The net force acting on the system is 98.15 N to the left, leading to an acceleration of approximately 0.9815 m/s². However, there is confusion regarding the application of mass in the net force equation, as both blocks influence the system's motion. Clarification is needed to ensure that the acceleration calculation accounts for the entire system rather than just the heavier block.
moondawg
Messages
45
Reaction score
0

Homework Statement


2 blocks connected by a cord passing over a small, frictionless pulley rest on frictionless planes. a.) Which way will the system move when the blocks are released from rest? b.) what is the acceleration of the blocks?
*** picture a triangle with a 100kg block over its left slops at angle of 30 degrees, and a 50kg block over its right slope with a 53.1 degree angle with a string through a fricitonless pulley connecting the 2 blocks together.


Homework Equations





The Attempt at a Solution


I found the forces exerted by each of the blocks by breaking up into components. I found the force for the 100 kg block to be 490N down the ramp and the force for the 50kg block to be 391.8N down the ramp so for a.) the system will movethe blokcs to the left(in the direction of 100kg block).
For b.) i found the net force to be 98.15N in the left direction then set 98.15=100a and found a to be .9815m/s2. I used 100kg as my mass bc that is the mass of block in which the direction of both blocks are going according to my data. But idk if that's right? please someone help me out?!
 
Physics news on Phys.org
98.15=100a
writing like this is like saying that the net force makes only 100 kg block to move, and 50 kg block isn't affected by it.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top