Pully with 2 blocks friction question

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SUMMARY

The problem involves determining the mass of block B required to initiate the movement of block A, which has a mass of 3.00 kg on a 30.0° inclined plane with a static friction coefficient of 0.400. The correct approach involves calculating the gravitational force acting on block A, which is given by F_gravity = m * g * sin(30°), and the frictional force, F_friction = μ * m * g * cos(30°). The tension in the string must overcome both the gravitational component and the frictional force to start block A sliding up the plane.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of static friction and its coefficient
  • Basic trigonometry for resolving forces
  • Familiarity with tension in strings and pulleys
NEXT STEPS
  • Calculate the gravitational force component acting on block A using F_gravity = m * g * sin(30°)
  • Determine the frictional force using F_friction = μ * m * g * cos(30°)
  • Learn about the dynamics of pulleys and tension in connected systems
  • Explore problems involving inclined planes and multiple blocks for further practice
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of inclined plane problems involving friction and tension in pulleys.

algar32
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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


http://ScrnSht.com/ftjykb
figure^^

The Attempt at a Solution


.4 x cos30 x 9.81 x 3kg = 10.2 =all forces down hill
ften = block b mass* 9.81
ftension must be equal to this to get the block started
10.2 = m x 9.81
m= 1kg
 
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algar32 said:
.4 x cos30 x 9.81 x 3kg = 10.2 =all forces down hill

This part is wrong. F[itex]_{gravity}[/itex] = mgsin30 and F[itex]_{friction}[/itex] = μmgcos30.
 
As pointed out by Tal444,
Tension will not be equal to 10.2

Tension is acting against two forces.
Force of friction and force of gravity (component along plane )
 

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