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. 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, and the tension T in the cord connecting A and C.
To calculate the acceleration of the system and the tensions in the cords, we need to consider the forces acting on each box.
Let's denote the acceleration of the system as "a." Since the system is released from rest, we know that the net force acting on the system is equal to the total mass of the system multiplied by its acceleration:
ΣF = (mA + mB + mC) * a
In this system, there are two tension forces and the force due to the gravitational pull acting on each box.
Now, let's analyze the forces acting on each box:
For box A:
- The tension force T1 is acting rightwards.
- The force of gravity pulling it downwards is mA * g, where g is the acceleration due to gravity.
For box B:
- The tension force T1 is acting leftwards.
- The tension force T is acting upwards.
- The force of gravity pulling it downwards is mB * g.
For box C:
- The tension force T is acting downwards.
- The force of gravity pulling it downwards is mC * g.
From the forces acting on each box, we can write down the equations of motion:
For box A: T1 - mA * g = mA * a
For box B: T1 - T - mB * g = mB * a
For box C: T - mC * g = mC * a
Now, let's solve the system of equations to find the values of acceleration and tensions.
We have three equations and three unknowns: a, T1, and T.
To solve the system of equations, you can use various methods, such as substitution or elimination. But to keep it simple, we'll use substitution.
We'll start by isolating T1 in terms of T in the second equation:
T1 = T + mB * g + mB * a
Now, substitute this value of T1 into the first equation:
T + mB * g + mB * a - mA * g = mA * a
Next, isolate T in terms of a in the third equation:
T = mC * g + mC * a
Now, substitute this value of T into the second equation:
T1 - (mC * g + mC * a) - mB * g = mB * a
After simplifying and rearranging the equations, you will have a quadratic equation in terms of a:
(a * (mB + mA + mC)) + (g * (mC - mA)) = (mB * g)
Simplify further:
a = (mB * g) / (mB + mA + mC)
This gives you the value of acceleration (a).
To find the tensions T1 and T, substitute the value of a back into the equations:
T1 = T + mB * g + mB * a
T = mC * g + mC * a
Now, substitute the value of a to calculate the tensions T1 and T.
I hope this explanation helps you understand the process of solving this problem.