Inextensible cords passing over frictionless pulleys

AI Thread Summary
The discussion centers on calculating the magnitude and direction of friction forces for two blocks connected by inextensible cords over frictionless pulleys, with coefficients of friction of 0.30 and 0.40 at points A and B, respectively. Users are attempting to determine the tension in the cords using trigonometric relationships, specifically the sine law, and are encouraged to draw Free Body Diagrams to visualize the forces acting on each block. The conversation emphasizes the importance of understanding Newton's laws of motion and the relationships between the blocks' accelerations and tensions. Participants express a desire for clarification on the formulas needed to solve the problem effectively. Overall, the thread highlights the complexities of the physics involved and the collaborative effort to grasp the concepts.
apprentice213
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Homework Statement


the blocks shown in the figure are connected by flexible, inextensible cords passing over frictionless pulleys. at A, the coefficient fiction is 0.30 while at B, it is 0.40. compute the magnitude and direction of the friction forces and each block.
am i right ? x( please help .. want to learn
untitled.jpg


Homework Equations



for block A:
teta(A)=tan^-1(o.30)
teta(A)=16.70
N 53.13 W
sine law:
T/sine53.13=300/sin36.87
T=400lbs

for block B:
teta(B)=tan^-1(0.40)
teta(B)=21.80
N 36.87 E
sine law:
T/sin36.87=200/sin53.13
T=150lbs

The Attempt at a Solution

 
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apprentice213 said:

Homework Statement


the blocks shown in the figure are connected by flexible, inextensible cords passing over frictionless pulleys. at A, the coefficient fiction is 0.30 while at B, it is 0.40. compute the magnitude and direction of the friction forces and each block.
am i right ? x( please help .. want to learn
untitled.jpg


Homework Equations



for block A:
teta(A)=tan^-1(o.30)
teta(A)=16.70
N 53.13 W
sine law:
T/sine53.13=300/sin36.87
T=400lbs

for block B:
teta(B)=tan^-1(0.40)
teta(B)=21.80
N 36.87 E
sine law:
T/sin36.87=200/sin53.13
T=150lbs

The Attempt at a Solution


I can't really see the set-up but can only advise that the usual calculation Friction = uR calculates the maximum possible friction. Perhaps in this case you don't need, and thus won't generate the maximum friction?
 


2.jpg


sorry :) this is the figure . my instructor said we must find the tension(that red line) .. what should be the possible formula?
 


then this the 1st thing i do

3-1.jpg


could it be solve by this figure ?

then followed by law of sine
 
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The force supporting each block in position comes in part by friction, and with a rope making up the difference. Friction may be acting up, or down, the slope, according to which way the block is tending to slide.
 
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what would be the possible formula sir? because i can't really understand this topic although i tried my best to solve .. please help thnx
 


anyway this topic is related to Newtons law of motion
 


You should draw Free Body Diagrams for both blocks, showing the forces acting. Given the mass of a block and the expression for the net force you should be able to write an expression for the acceleration of each block. These expressions will contain the (as yet) unknown values of the tensions in the two ropes.

Next examine the original figure and determine the relationships between the two accelerations and the two tensions that will be enforced by the pulley setup. You should then have enough information to solve for the tensions.
 


thnx sir :)
 
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