 Annette Arroyo

2020-10-18

Three identical blocks connected by ideal strings are being pulled along a horizontal frictionless surface by a horizontal force $\stackrel{\to }{F}$ The magnitude of the tension in the string between blocks B and C is T=3.00N. Each block has mass m=0.400kg.
What is the magnitude F of the force?
What is the tension in the string between block A and block B? Ezra Herbert

The magnitude of F of the force
T=ma
$3.00N=0.400kg\cdot a$
thus

However, this only represents the overall acceleration; for each separate block, the

F can now be find by using
$a=\frac{F}{3m}$ and solving for F
$F=a\cdot 3m$
$F=4.5N$
For the Tension between block A and B, all we do is
T=ma, knowing that  for each block. Jeffrey Jordon

${T}_{2}=3N$

M=0.4 kg

${T}_{1}=0.4a$  -(1)

${T}_{2}-{T}_{1}=0.4a$ -(2)

$F-{T}_{2}=0.4a$ -(3)

$a=3.75m/{s}^{2},{T}_{AB}=1.5N$ Vasquez

Result:
- The magnitude of the force $F$ is $3.00\phantom{\rule{0.167em}{0ex}}\text{N}$.
- The tension in the string between block A and block B is $3.00\phantom{\rule{0.167em}{0ex}}\text{N}$.
Solution:
Let's denote the magnitude of the force as $F$.
Since the three blocks are connected by ideal strings, the tension in the string between blocks B and C is equal to the force applied on block C. Therefore, the tension $T$ is equal to the force applied on block C, which can be expressed as:
$T=F$
Given that $T=3.00\phantom{\rule{0.167em}{0ex}}\text{N}$, we have:
$F=3.00\phantom{\rule{0.167em}{0ex}}\text{N}$
Now, to find the tension in the string between block A and block B, we need to consider the motion of the blocks as a system.
The force applied on block A is equal to the sum of the tension in the string between A and B and the tension in the string between B and C. Therefore, the force applied on block A can be expressed as:
${F}_{A}=T+T$
Since the masses of all three blocks are identical, the acceleration of each block is the same. Applying Newton's second law to each block, we have:
${F}_{A}=m·a$
Where $m$ is the mass of each block and $a$ is the common acceleration of the blocks.
Since the masses of the blocks are given as $m=0.400\phantom{\rule{0.167em}{0ex}}\text{kg}$, we substitute this value into the equation:
${F}_{A}=0.400\phantom{\rule{0.167em}{0ex}}\text{kg}·a$
Since the force applied on block A is equal to the force $F$, we can rewrite the equation as:
$F=0.400\phantom{\rule{0.167em}{0ex}}\text{kg}·a$
Now we can solve for the acceleration $a$. Dividing both sides of the equation by $0.400\phantom{\rule{0.167em}{0ex}}\text{kg}$, we get:
$a=\frac{F}{0.400\phantom{\rule{0.167em}{0ex}}\text{kg}}$
Since $F=3.00\phantom{\rule{0.167em}{0ex}}\text{N}$, we substitute this value into the equation:
$a=\frac{3.00\phantom{\rule{0.167em}{0ex}}\text{N}}{0.400\phantom{\rule{0.167em}{0ex}}\text{kg}}$
Simplifying the expression, we find:
$a=7.50\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}$
Finally, to find the tension in the string between block A and block B, we use the equation:
${T}_{AB}={F}_{A}-T$
Substituting the known values, we have:
${T}_{AB}=2T-T$
${T}_{AB}=T$
Therefore, the tension in the string between block A and block B is also equal to $3.00\phantom{\rule{0.167em}{0ex}}\text{N}$. Don Sumner

Let's denote the magnitude of the force $\stackrel{\to }{F}$ as $F$, the tension in the string between block A and B as ${T}_{AB}$, and the mass of each block as $m=0.400\phantom{\rule{0.167em}{0ex}}\text{kg}$.
Using Newton's second law, we know that the net force acting on an object is equal to its mass multiplied by its acceleration. Since the blocks are connected by ideal strings, they will all have the same acceleration.
First, let's analyze the forces acting on block A:
- The tension ${T}_{AB}$ in the string is directed towards the right.
- There is no other force acting on block A horizontally.
Therefore, we can write the equation of motion for block A as:

Next, let's analyze the forces acting on block B:
- The tension ${T}_{AB}$ in the string is directed towards the left.
- The tension $T$ in the string between blocks B and C is directed towards the right.
Therefore, we can write the equation of motion for block B as:

Finally, let's analyze the forces acting on block C:
- The tension $T$ in the string between blocks B and C is directed towards the left.
- There is no other force acting on block C horizontally.
Therefore, we can write the equation of motion for block C as:

Since the blocks are connected and experiencing the same acceleration, we can equate the equations of motion for each block:
${T}_{AB}-F={T}_{AB}-T=T$
Given that $T=3.00\phantom{\rule{0.167em}{0ex}}\text{N}$, we can solve for $F$ and ${T}_{AB}$:
$F={T}_{AB}-T=3.00\phantom{\rule{0.167em}{0ex}}\text{N}$
${T}_{AB}=T+T=3.00\phantom{\rule{0.167em}{0ex}}\text{N}+3.00\phantom{\rule{0.167em}{0ex}}\text{N}=6.00\phantom{\rule{0.167em}{0ex}}\text{N}$
Hence, the magnitude of the force $F$ is $3.00\phantom{\rule{0.167em}{0ex}}\text{N}$, and the tension in the string between block A and B is $6.00\phantom{\rule{0.167em}{0ex}}\text{N}$. nick1337

Step 1:
For block A, there are two forces acting on it: the tension ${T}_{AB}$ in the string between A and B, and the force $\stackrel{\to }{F}$ pulling the blocks. Since block A is not accelerating in the vertical direction, the vertical components of these forces must cancel each other out.
Let's consider the horizontal forces acting on block A. The only horizontal force is the tension ${T}_{AB}$ in the string between A and B, which is directed to the right. Therefore, we can write the equation of motion for block A as:
${T}_{AB}-F=0$
Since the magnitude of ${T}_{AB}$ is not given, we need to solve for it.
For block B, there are three forces acting on it: the tension ${T}_{AB}$ in the string between A and B, the tension ${T}_{BC}$ in the string between B and C, and the force $\stackrel{\to }{F}$ pulling the blocks. Again, considering the vertical forces, we can conclude that the tension ${T}_{BC}$ and the vertical component of ${T}_{AB}$ must cancel each other out.
Let's focus on the horizontal forces acting on block B. The tension ${T}_{AB}$ is directed to the left, and the tension ${T}_{BC}$ is directed to the right. We can write the equation of motion for block B as:
${T}_{BC}-{T}_{AB}=0$
Finally, for block C, the only horizontal force acting on it is the tension ${T}_{BC}$, which is directed to the left. We can write the equation of motion for block C as:
${T}_{BC}-F=0$
Step 2:
Now, we can solve these equations simultaneously to find the values of $F$ and ${T}_{AB}$.
From the equation for block B, we can see that ${T}_{AB}={T}_{BC}=T=3.00\phantom{\rule{0.167em}{0ex}}\text{N}$.
Substituting this value into the equation for block A, we have:
$3.00\phantom{\rule{0.167em}{0ex}}\text{N}-F=0$
Simplifying, we find:
$F=3.00\phantom{\rule{0.167em}{0ex}}\text{N}$
Therefore, the magnitude of the force $\stackrel{\to }{F}$ is $3.00\phantom{\rule{0.167em}{0ex}}\text{N}$.
To find the tension ${T}_{AB}$, we can substitute the known values into the equation for block B:
${T}_{BC}-{T}_{AB}=0$
${T}_{AB}={T}_{BC}=T=3.00\phantom{\rule{0.167em}{0ex}}\text{N}$
Therefore, the tension in the string between block A and block B is ${\text{T}}_{\text{AB}}=3.00\phantom{\rule{0.167em}{0ex}}\text{N}$.

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