Consider the following system with a 2 kg mass sitting on a table, connected to both 1 kg and 3 kg masses by strings which go over two separate pulleys and are pulling on the 2 kg mass in opposite directions. What static friction coefficient would be required between the 2 kg mass and the table to stop the masses from accelerating? If the masses are in motion, what kinetic friction coefficient would be required between the 2 kg mass and the table to ensure the masses move at a constant velocity? Find the tension in both pieces of string for the case of zero friction
net force=2g-1g-2g*mu=0
solve for muStatic
same equation for moving, only mu is now muKinetic
Tension in both strings=weight of the mass hanging over.
net force=3g-1g-2g*mu=0
solve for muStatic
same equation for moving, only mu is now muKinetic
Tension in both strings=weight of the mass hanging over.
See correction above, I typed in the hanging mass as 2 and 1, when they should be 3kg and 1kg
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To determine the static friction coefficient required to stop the masses from accelerating, we need to consider the forces acting on the 2 kg mass.
When the system is at rest, the net force on the 2 kg mass must be zero. This means that the tension in the string attached to the 1 kg mass, T1, must be equal to the tension in the string attached to the 3 kg mass, T2. Since these tensions are pulling in opposite directions, the static friction force between the 2 kg mass and the table must also be equal to the tension in the string attached to the 1 kg mass.
We can write the equation for the forces acting on the 2 kg mass in the horizontal direction:
T2 - T1 - f_static = 0
Where T2 is the tension in the string attached to the 3 kg mass, T1 is the tension in the string attached to the 1 kg mass, and f_static is the static friction force between the 2 kg mass and the table.
Since T2 = T1, the equation becomes:
2T1 - f_static = 0
Therefore, we can conclude that the static friction force between the 2 kg mass and the table must be equal to twice the tension in either string.
If the masses are in motion and we want them to move at a constant velocity, we need to consider the kinetic friction coefficient between the 2 kg mass and the table.
In this case, the net force on the 2 kg mass is zero, meaning the tension in the string attached to the 1 kg mass must be equal to the tension in the string attached to the 3 kg mass plus the kinetic friction force between the 2 kg mass and the table.
We can write the equation for the forces acting on the 2 kg mass in the horizontal direction:
T2 - T1 - f_kinetic = 0
Where T2 is the tension in the string attached to the 3 kg mass, T1 is the tension in the string attached to the 1 kg mass, and f_kinetic is the kinetic friction force between the 2 kg mass and the table.
Since T2 = T1, the equation becomes:
2T1 - f_kinetic = 0
Therefore, we can conclude that the kinetic friction force between the 2 kg mass and the table must be equal to twice the tension in either string.
Now, let's find the tension in both pieces of string for the case of zero friction. If there is no friction between the 2 kg mass and the table, the net force will be zero, and the tension in each string will be equal to the gravitational force acting on the mass it is connected to.
For the string attached to the 1 kg mass:
T1 = m1 * g
Where m1 is the mass of the 1 kg mass and g is the acceleration due to gravity.
For the string attached to the 3 kg mass:
T2 = m2 * g
Where m2 is the mass of the 3 kg mass and g is the acceleration due to gravity.
Therefore, for the case of zero friction, the tension in each string will be equal to the gravitational force acting on the corresponding mass.