Two boxes connected to a string over an ideal pulley?

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The discussion focuses on a physics problem involving two blocks connected by a string over a pulley, where block A, with a mass of 3.00 kg, is on a 30° inclined plane. The goal is to determine the mass of block B needed to initiate the sliding of block A up the plane, considering static friction with a coefficient of 0.400. Key forces acting on block A include static friction, normal force, gravity, and tension. Participants analyze the equations of motion for both blocks, specifically the relationship between friction, gravity, and tension. The calculated mass for block B to start moving block A is approximately 3.198 kg.
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Homework Statement


Two blocks are connected by a string that goes over an ideal pulley as shown in the figure and pulls on block A parallel to the surface of the plane. Block A has a mass of 3.00 kg and can slide along a rough plane inclined 30.0° to the horizontal. The coefficient of static friction between block A and the plane is 0.400. What mass should block B have in order to start block A sliding up the plane?


Homework Equations


box A- sum of x & sum of y
box B- sum of y
plug into each other



The Attempt at a Solution


m2=(ms*m1*sin(theta)+m1*cos(theta)
=3.198


Thanks.
 
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First of all. What are all the forces acting upon block A?
 
tal444 said:
First of all. What are all the forces acting upon block A?

The forces acting on block A are static friction, fnormal, gravity, and ftension.
 
Yes. So the friction force and force of gravity are preventing the block from sliding up the ramp. What's next?
 
tal444 said:
Yes. So the friction force and force of gravity are preventing the block from sliding up the ramp. What's next?

ffriction = -.4*9.81*3
fgrav= -sin30*9.81*3
ftension =m*9.81

ffriction +fgrav = ftension

Not sure which part i messed up
 
Last edited:
algar32 said:
ffriction = -.4*9.81*3

Friction force is equal to μF_{normal}.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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