How Does the Tension Equation Differ with Inclined Planes and Multiple Masses?

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The discussion revolves around the tension equation for a system involving an inclined plane and two masses. The original poster encountered a tension formula in their textbook that differed from their understanding, leading to confusion. They initially believed tension equaled the sum of forces applied to the rope but later realized that analyzing forces on each mass using Newton's second law was necessary. The key takeaway is that the correct approach involves considering acceleration rather than just gravitational forces when calculating tension. Ultimately, the poster corrected their mistake and understood the proper method for solving the problem.
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[SOLVED] Tension Equation Question

Hi, this is my first post here.

i came across an equation for tension in my textbook that i have never seen before and wanted to ask if someone could explain it to me.

Tension = g * (m^{1} * m^{2}) / (m^{1} + m^{2}) * (1 + sin\Theta)


i always thought Tension is equal to the sum of the forces applied to the rope.


the context is following problem:

a block of mass m_{1} is at rest on an inclined plane at \Theta degrees with the horizontal. it is connected with a block of mass m_{2} that is hanging of the inclined plane hrough a massless rope.


so the way i thought about tension it would just be

Tension = m_{1} * g * sin\Theta + m_{2} * g

but that gives me a different answer and i don't know what's wrong.

so any help is appreciated.
 
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ThatGermanDude said:
i always thought Tension is equal to the sum of the forces applied to the rope.
The tension is the force exerted on each end of the rope (and exerted by each end of the rope). Note that for a massless rope, the force at each end is the same--the tension is the same throughout the rope.

so the way i thought about tension it would just be

Tension = m_{1} * g * sin\Theta + m_{2} * g
Realize the the weights are forces that act directly on the masses, not the rope. To find the tension in the rope, analyze the forces on each mass: On m1 there is gravity and the rope tension; on m2 there is also gravity and rope tension. Apply Newton's 2nd law to each mass and solve for the tension.
 
Draw a force diagram for the two blocks with the blocks having an acceleration 'a' and a tension 'T'. Then actually solve for T and a. Assuming it's the sum of the two gravitational forces is just jumping to conclusions.
 
i did all that and then end up with the equation i stated before.
in this case the only acting forces are th weights since the slope is frictionless.
 
Specify the forces acting on each mass and the resulting equations you got from applying Newton's 2nd law.
 
i figured it out. the mistake i made was using gravity instead of acceleration, like i should have, to find tension.

thanks
 
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}...

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