two masses are attached to both ends of a flexible cord passing through a friction less pulley. if the system is at rest, find the tensions at object A and object B. Object A = Object B. A=60 grams.
Tension? Pulling one way, mg= 9.8*.06
and because it is not moving, that is the tension
0.588 N
To find the tensions at objects A and B in the given system, we need to consider the equilibrium condition. Since the system is at rest, the net force acting on each object must be zero.
Let's assume that the mass attached to object A is denoted as mA, and the mass attached to object B is denoted as mB. In this case, the mass mA is given as 60 grams.
In a simple setup where the cord passes through a frictionless pulley, the tension in the cord on both sides of the pulley is the same. This means that the tension on object A is equal to the tension on object B.
Next, we need to determine the force acting on each object. The force acting on an object is given by the product of its mass and acceleration. In this case, since the system is at rest, the acceleration is zero, and therefore, the force acting on each object is zero.
Now, using Newton's second law (F = m * a), we can calculate the force acting on each object:
For object A:
Force on object A = mass of object A * acceleration
Force on object A = mA * 0 (as the system is at rest)
Force on object A = 0
For object B:
Force on object B = mass of object B * acceleration
Force on object B = mB * 0 (as the system is at rest)
Force on object B = 0
Since both forces are zero, the tension at object A and the tension at object B are also zero.
Therefore, the tensions at object A and object B in this system, when it is at rest, are zero.