Three boxes are connected by cords, one of which (mA) wraps over a pulley having negligible friction on its axle and negligible mass and two boxes (mB & mC) are hanging. The masses are mA = 22.0 kg, mB = 40.0 kg, mC = 16.0 kg.
-When the assembly is released from rest, what is the tension in the cord that connects boxes B and C?
-How far (in meters) does box A move in the first 0.250 s (assuming it does not reach the pulley)?
Write the force equation clockwise being positive.
Ma*g-Mb*g-Mc*g=totalmass*acceleration.
where total mass is the sum of the three masses. Solve for acceleration.
Tension that connects to B is the same as the tension that is connected to A.
Ma*g-Tension=totalmass*acceleration
displacement= 1/2 acceleration*time^2
To find the tension in the cord that connects boxes B and C, we need to first calculate the acceleration of the system. We can use the force equation you provided:
Ma*g - Mb*g - Mc*g = totalmass * acceleration
Plugging in the given values:
Ma = 22.0 kg, Mb = 40.0 kg, Mc = 16.0 kg
We also know that acceleration = 0 initially because the assembly is released from rest, so the equation becomes:
22.0 kg * 9.8 m/s^2 - 40.0 kg * 9.8 m/s^2 - 16.0 kg * 9.8 m/s^2 = (22.0 kg + 40.0 kg + 16.0 kg) * acceleration
Simplifying this equation:
215.6 N - 392.0 N - 156.8 N = 78.0 kg * acceleration
-333.2 N = 78.0 kg * acceleration
Solving for acceleration:
acceleration = -333.2 N / 78.0 kg
acceleration ≈ -4.275 m/s^2 (the negative sign indicates the direction of motion)
Since the tension that connects to B is the same as the tension that is connected to A, we can now find the tension in the cord between B and C using the following equation:
Ma * g - Tension = totalmass * acceleration
Plugging in the given values, we get:
(22.0 kg) * (9.8 m/s^2) - Tension = (22.0 kg + 40.0 kg + 16.0 kg) * (-4.275 m/s^2)
Simplifying this equation:
215.6 N - Tension = (78.0 kg) * (-4.275 m/s^2)
Rearranging the equation to solve for Tension:
Tension = 215.6 N - 78.0 kg * (-4.275 m/s^2)
Tension ≈ 541.5 N
So, the tension in the cord that connects boxes B and C is approximately 541.5 N.
To find how far box A moves in the first 0.250 s, we can use the equation of motion for constant acceleration:
displacement = (1/2) * acceleration * time^2
Plugging in the values:
displacement = (1/2) * (-4.275 m/s^2) * (0.250 s)^2
Simplifying this equation:
displacement ≈ -0.267 m
Therefore, box A moves approximately -0.267 meters in the first 0.250 seconds (assuming it does not reach the pulley). The negative sign indicates that it moves in the opposite direction of the positive reference point.