Determine the magnitude of the tension in the clotheline

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To determine the tension in the clothesline with a pulley, the setup involves a right triangle where the horizontal distance is 5 m and the vertical drop is 0.4 m. Using the relationship 2*T*cos(θ) = mg, where mg is the weight of the pulley (294 N), the angle θ can be calculated as cos(θ) = 0.4/5. This leads to the correct tension T being approximately 1800 N, which is half of the initial incorrect calculation of 3686.74 N. The discussion also briefly touches on a separate problem involving forces between two burger patties, emphasizing the need to apply Newton's second law to find the force exerted by one patty on the other. Understanding these principles is crucial for solving similar physics problems effectively.
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Homework Statement


a single clotheline is attached to two poles 10 m apart. A pulley with a mass of 294N/Fg rolls down to the middle of the line. the middle is 0.4 m below each end. Determine the magnitude of the tension in the clotheline.


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The Attempt at a Solution


I have made a right angle triangle and then used the ratios 0.4 and 5 m to find the force of the adjacent side of the triangle. then i use the given 294N and the adjacent 3675N in the pythagorean theorem. i get 3686.74N but the answer is 1800N rounded. which is half of my answer.
 
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Hi stateofdogma, welcome to PF.
Since the pulley is free to move on the cloth line, tension in the two segments of the cloth line must be the same. It θ is the angle between the vertical and the segment, then
2*T*cosθ = mg.
Here cosθ = 0.4/5. mg is given. find T.
 
Thanks for the homework help.
 
this is a question which i figured out on my second try. two burger patties, in contact with each other are being pushed across a grill. the masses are 113g and 139g, a force of 0.0538N is used on the larger patty. I figured out the acceleration was 0.213m/s^2 by dividing the force by the masses. But the second part asks for the magnitude of the force exerted by each of the burger on the other. the answer is the acceleration multiplied by the 113g which is 0.0241N. the problem is i don't understand why
thanks in advance
 
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Thread 'Variable mass system : water sprayed into a moving container'

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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