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John Rennie
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Blocks

I'm guessing your FBD looked something like this, where $F_1$ is the external force you apply. I'm assuming here that the top and bottom block don't slide relative to each other, so the forces at their junction ($F_2$) are equal and opposite.

The net force on the top block is the force you apply, $F_1$, minus the frictional force the bottom block applies to the top block, $F_2$:

$$ F_{top} = F_1 - F_2 $$

Because $F_1 > F_2$ the net force $F_{top} > 0$ and the top block accelerates.

Response to comment:

If the two blocks don't slide relative to each other then their accelerations must be the same so:

$$ \frac{F_{top}}{m_{top}} = \frac{F_{bottom}}{m_{bottom}} $$

We know that $F_{top} = F_1 - F_2$ and $F_{bottom} = F_2$, so:

$$ \frac{F_1 - F_2}{m_{top}} = \frac{F_2}{m_{bottom}} $$

and a quick rearrangement gives:

$$ F_1 = F_2 \frac{m_{top} + m_{bottom}}{m_{bottom}} $$

and since $m_{top} + m_{bottom} > m_{bottom}$ this means $F_1 > F_2$.

Basically $F_2$ is only accelerating the bottom block while $F_1$ is accelerating both blocks, so $F_1$ has to be greater than $F_2$.

Blocks

I'm guessing your FBD looked something like this, where $F_1$ is the external force you apply. I'm assuming here that the top and bottom block don't slide relative to each other, so the forces at their junction ($F_2$) are equal and opposite.

The net force on the top block is the force you apply, $F_1$, minus the frictional force the bottom block applies to the top block, $F_2$:

$$ F_{top} = F_1 - F_2 $$

Because $F_1 > F_2$ the net force $F_{top} > 0$ and the top block accelerates.

Blocks

I'm guessing your FBD looked something like this, where $F_1$ is the external force you apply. I'm assuming here that the top and bottom block don't slide relative to each other, so the forces at their junction ($F_2$) are equal and opposite.

The net force on the top block is the force you apply, $F_1$, minus the frictional force the bottom block applies to the top block, $F_2$:

$$ F_{top} = F_1 - F_2 $$

Because $F_1 > F_2$ the net force $F_{top} > 0$ and the top block accelerates.

Response to comment:

If the two blocks don't slide relative to each other then their accelerations must be the same so:

$$ \frac{F_{top}}{m_{top}} = \frac{F_{bottom}}{m_{bottom}} $$

We know that $F_{top} = F_1 - F_2$ and $F_{bottom} = F_2$, so:

$$ \frac{F_1 - F_2}{m_{top}} = \frac{F_2}{m_{bottom}} $$

and a quick rearrangement gives:

$$ F_1 = F_2 \frac{m_{top} + m_{bottom}}{m_{bottom}} $$

and since $m_{top} + m_{bottom} > m_{bottom}$ this means $F_1 > F_2$.

Basically $F_2$ is only accelerating the bottom block while $F_1$ is accelerating both blocks, so $F_1$ has to be greater than $F_2$.

Source Link
John Rennie
  • 362.6k
  • 132
  • 780
  • 1.1k

Blocks

I'm guessing your FBD looked something like this, where $F_1$ is the external force you apply. I'm assuming here that the top and bottom block don't slide relative to each other, so the forces at their junction ($F_2$) are equal and opposite.

The net force on the top block is the force you apply, $F_1$, minus the frictional force the bottom block applies to the top block, $F_2$:

$$ F_{top} = F_1 - F_2 $$

Because $F_1 > F_2$ the net force $F_{top} > 0$ and the top block accelerates.