Shows three boxes connected by cords, one of which wraps over a pulley having negligible friction on its axle and negligible mass. The mass of the boxes are mA = 5 kg, mB = 2 kg, and mC = 3kg. Given that there is no friction between box A and box B with the surface of the plane.(the plane is mass A and mass B which are together put with some horizontal distance and mass A and mass B are perpendicular to mass C which is hanging. When the system is released from rest, Calculate (a) the acceleration of the system of mass, (b) the tension T1 in the cord connecting A and B, (c) the tension T in the cord connecting A and C.
To solve this problem, we need to apply Newton's second law of motion to each box in the system. Let's break down the problem into parts:
(a) Acceleration of the system:
We can start by considering the forces acting on each object separately:
Box A:
- Tension force T1 in the direction of motion.
- The weight of mA acting downward.
Box B:
- Tension force T1 in the direction of motion.
- The weight of mB acting downward.
Box C:
- Tension force T in the upward direction.
- The weight of mC acting downward.
Now let's write the equations of motion for each box using Newton's second law (F = ma) and considering the force components along the x-axis:
For Box A:
T1 - mA * g = mA * a.
For Box B:
T1 - mB * g = mB * a.
For Box C:
T - mC * g = mC * a.
Since the system is connected and each box has the same acceleration (a), we can sum up the equations of motion for all three boxes:
T1 - mA * g - mB * g - mC * g = (mA + mB + mC) * a.
Substituting the given values:
T1 - 5 * 9.8 - 2 * 9.8 - 3 * 9.8 = (5 + 2 + 3) * a.
Now we can solve for a:
T1 - 98 - 19.6 - 29.4 = 10 * a,
T1 - 147 = 10 * a.
(b) Tension T1 in the cord connecting A and B:
Using the equation found in part (a), we can substitute the value of a into the equation:
T1 - 147 = 10 * a.
Substituting the value of a = (T1 - 147) / 10 into any of the previous equations (e.g., for Box A, T1 - 5 * 9.8 = 5 * a), we can solve for T1.
(c) Tension T in the cord connecting A and C:
To find the tension in the cord connecting A and C, we need to consider the forces acting on Box A. There are two tensions involved: T1 pulling to the right and T pulling upwards.
Using the equation for Box A:
T1 - mA * g = mA * a.
Substituting the given values, we can solve for T:
T - 5 * 9.8 = 5 * a.
Once you have the value of a from part (a), you can substitute it into this equation to find T.
Remember, in order to obtain accurate values, it is important to use the correct units (e.g., m/s^2 for acceleration, N for force, kg for mass, etc.) throughout the calculations.