Suppose two boxes on a frictionless table are connected by a heavy cord of mass 1.0 kg. Calculate the acceleration (magnitude) of each box and the tension (magnitude) at each end of the cored, using the free-body diagrams shown in Figure 4-53. Assume FP = 44.0 N, ma = 11.0 kg, and mb = 13.5 kg, and ignore sagging of the cord. Compare your results to Example 4-12 and Figure 4-22.
mA acceleration ____ m/s2
mB acceleration ____ m/s2
FTA ____ N
FTB ____ N
If you treat the whole system as one solid thing, you will get the answer.
FP= mtotal x acceleration
44= (11+13.5+1)a
44=25.5a
a=1.73m/s^2
Use this acceleration value to find FTA and FTB using Newton's 2nd Law.
FBT=13 x 1.73
FBT= 22.5 N
(FBT and FTB are equal in magnitude)
and
44-FAT=11 x 1.73
44-FAT=19.03
FAT=44-19.03
FAT=25.0 N
Why did the cord go to the circus? Because it wanted to show off its ac-cord-ion skills! Now, let's dive into the problem.
To find the acceleration of each box, we can use the equation:
Force = mass * acceleration
For box A:
FP - FTA = mA * aA
Rearranging the equation, we get:
FTA = FP - mA * aA
For box B:
FTB - FP = mB * aB
Rearranging the equation, we get:
FTB = FP + mB * aB
Now, comparing the equations to Example 4-12 and Figure 4-22, we can see that the only difference is the masses and forces involved. So, let's substitute the given values into the equations and solve:
FTA = 44.0 N - 11.0 kg * aA
FTB = 44.0 N + 13.5 kg * aB
Next, we need to consider the fact that the cord connecting the boxes has a mass of 1.0 kg. We have to account for its weight in our calculations. The force acting on the cord (FCord) due to its weight is given by:
FCord = mCord * g
Substituting the value, we have:
FCord = 1.0 kg * 9.8 m/s^2
FCord = 9.8 N
Since the cord is massless, the force acting on it is equal at both ends:
FTA = FTB = FCord
Now, we can set up two equations using the equations we derived earlier:
FTA = 44.0 N - 11.0 kg * aA
FTB = 44.0 N + 13.5 kg * aB
But since we have a third equation, we know FTA = FTB = FCord, we can substitute these values into the other two equations:
44.0 N - 11.0 kg * aA = 9.8 N
44.0 N + 13.5 kg * aB = 9.8 N
Now, solving these equations will give us the acceleration values for each box:
-11.0 kg * aA = 9.8 N - 44.0 N
aA = (9.8 N - 44.0 N) / -11.0 kg
aA ≈ -3.0 m/s^2 (Note: since this is negative, it means the acceleration is opposite to the direction of the applied force.)
13.5 kg * aB = 9.8 N - 44.0 N
aB = (9.8 N - 44.0 N) / 13.5 kg
aB ≈ -3.0 m/s^2 (Note: again, the negative sign implies it is in the opposite direction.)
Finally, let's calculate the tension at each end of the cord:
FTA = FCord = 9.8 N
FTB = FCord = 9.8 N
So, the answers are:
mA acceleration ≈ -3.0 m/s^2
mB acceleration ≈ -3.0 m/s^2
FTA ≈ 9.8 N
FTB ≈ 9.8 N
Hope this helps with your physics problem! Remember, physics can be a real tightrope act!
To solve this problem, we can apply Newton's second law of motion to each box separately.
For box A with mass mA = 11.0 kg:
The forces acting on box A are the tension force FT from the cord and the applied force FP = 44.0 N. The acceleration of box A is given by:
ma = FT / mA
ma = FT / 11.0 kg
For box B with mass mB = 13.5 kg:
The forces acting on box B are the tension force FT from the cord and the gravitational force Fg. The acceleration of box B is given by:
mB = FT / mB
mB = FT / 13.5 kg
To find FT, we can consider the system as a whole. Since the cord is inextensible and has mass, the tension force is the same throughout the cord. We can equate the magnitude of the tension force FT at each end of the cord.
FTA = FTB = FT
Now, we need to solve the system of equations. Let's start by solving for FT:
From the equation for box A:
ma = FT / 11.0 kg
FT = ma * 11.0 kg
Substituting the given values:
FT = (11.0 kg) * (ma)
From the equation for box B:
mB = FT / 13.5 kg
FT = mB * 13.5 kg
Substituting the given values:
FT = (13.5 kg) * (mB)
Since FT is the same in both equations, we can set them equal to each other:
ma * 11.0 kg = mB * 13.5 kg
Now, let's solve for ma and mB:
ma = (mB * 13.5 kg) / 11.0 kg
mB * 13.5 kg = ma * 11.0 kg
Dividing both sides by (11.0 kg * 13.5 kg):
ma = mB * (13.5 kg) / (11.0 kg * 13.5 kg)
ma = mB / 11.0 kg
We know that:
ma + mB = 11.0 kg + 13.5 kg = 24.5 kg
Substituting this into the equation for ma:
ma = (24.5 kg - ma) / 11.0 kg
Multiplying both sides by 11.0 kg:
11.0 kg * ma = 24.5 kg - ma
Simplifying:
12.0 kg * ma = 24.5 kg
Dividing both sides by 12.0 kg:
ma = 24.5 kg / 12.0 kg
ma = 2.04 m/s^2
Since ma + mB = 24.5 kg, we can substitute ma in the equation:
2.04 m/s^2 + mB = 24.5 kg
Subtracting 2.04 m/s^2 from both sides:
mB = 24.5 kg - 2.04 m/s^2
mB = 22.46 kg
Now, we can calculate the tension force FT:
FT = ma * 11.0 kg
FT = (2.04 m/s^2) * (11.0 kg)
FT = 22.4 N
Finally, the tension forces at each end of the cord are:
FTA = FT = 22.4 N
FTB = FT = 22.4 N
Therefore, the values are:
mA acceleration = 2.04 m/s^2
mB acceleration = 22.46 kg
FTA = 22.4 N
FTB = 22.4 N
These results should be compared to Example 4-12 and Figure 4-22 for validation.
To calculate the acceleration of each box and the tension at each end of the cord, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
Let's start by calculating the acceleration of box A (ma):
We are given that FP (the force pulling box B to the right) is 44.0 N.
Since there is no friction on the table, the only horizontal force acting on box A is the tension in the cord, T. The tension in the cord will be the same on both sides of the cord due to its mass being negligible. Therefore, the tension on box A is also T.
Using Newton's second law for box A:
FP - T = ma * a (equation 1)
Now let's calculate the acceleration of box B (mb):
The only horizontal force acting on box B is the tension in the cord, T. The tension in the cord will be the same on both sides of the cord, so the tension on box B is also T.
Using Newton's second law for box B:
T = mb * b (equation 2)
Now, we can solve these equations simultaneously to find the acceleration of each box and the tension at each end of the cord.
Substituting equation 2 into equation 1:
FP - mb * b = ma * a
Substituting the given values:
44.0 N - (13.5 kg * b) = 11.0 kg * a (equation 3)
To solve this equation, we need to know the relationship between the acceleration of box A (a) and the acceleration of box B (b). According to the diagram, the two boxes are connected by a heavy cord. Since the cord is considered massless, the acceleration of the two boxes will be the same. Therefore:
a = b (equation 4)
Substituting equation 4 into equation 3:
44.0 N - (13.5 kg * a) = 11.0 kg * a
Rearranging the equation:
44.0 N = 24.5 kg * a
Solving for a:
a = 44.0 N / 24.5 kg
Simplifying the calculation:
a ≈ 1.80 m/s^2
Now that we know the acceleration of box A, we can determine the tension at each end of the cord.
Using equation 2:
T = mb * b
Substituting the given values:
T = 13.5 kg * 1.80 m/s^2
Calculating T:
T ≈ 24.3 N
Therefore, the answers are as follows:
mA acceleration: 1.80 m/s^2
mB acceleration: 1.80 m/s^2
FTA (tension at end A): 24.3 N
FTB (tension at end B): 24.3 N