Difference between applied tension and tension due to weight?

AI Thread Summary
The discussion centers on understanding the difference between applied tension and tension due to weight in a physics problem involving a block on a table and a hanging mass. In Case I, a 50 N force is applied directly to the string, resulting in a tension of 50 N. In Case II, a 5 kg block is attached, leading to a calculated tension of 40 N due to the presence of additional mass and friction. The equations reveal that the acceleration is lower in Case II because the system's net force is reduced by the weight of the block and frictional forces. This illustrates how the dynamics of the system affect the tension in the string.
Petrikovski
im trying to do this AP problem that involves a block on a table attached to a string which hangs over a pulley. in Case I 50 N is applied to the string and in Case II a 5 kg block is attached. the tension in Case I is 50, but according to my packet in case II the tension in Case II is 40. why would 50 N applied directly be more tension than Fw = 5x10 = 50 N?
 
Physics news on Phys.org
btw the block on top is 10kg and there is a coefficient of friction of .2
 
Because in the second case there is more mass so there's less acceleration:
mg - Fs = (M+m)a
50 - 20 = 15a
a = 2
Now looking at the 5kg weight:
50 - T = 5a
T = 50 - 10 = 40
 
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...
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}...
Back
Top