Question

A horizontal force is applied on a uniform rod of length L kept on a frictionless surface. Find the tension in rod at a distance 'x' from the end where force is applied.

Considering rod as a system, we find acceleration of rod a = $\frac{F}{M}$

now draw F.B.D. of rod having length 'x' as shown in figure. Using Newton's second law F-T = $(\frac{M}{L})$x.a $\Rightarrow$ T = F - $\frac{M}{L}$x.$\frac{F}{M}$ $\Rightarrow$ T = F(1-$\frac{x}{L}$).

Answer 1

T = F(1 - x/L)

Answer 2

Solution: The rod has uniform mass distribution. Let the total mass be $M$ and length be $L$. A horizontal force $F$ is applied at one end. The rod is on a frictionless surface, so the net force is $F$.

1. Acceleration of the rod:
   Using Newton's second law for the entire rod:
   $F_{net} = Ma$
   $F = Ma$
   $a = \frac{F}{M}$

2. Free-body diagram of a section:
   Consider a section of the rod of length $x$ from the end where the force $F$ is applied.
   The mass of this section is $m_x = \left(\frac{M}{L}\right)x$.
   The forces acting on this section are:

   • The applied force $F$ at the left end.
   • The tension $T$ exerted by the remaining part of the rod at the cut (at distance $x$). This tension opposes the motion of this section relative to the remaining part, or more accurately, it's the force transmitted through the rod. Since the rod is accelerating to the right, the part to the right of the cut pulls the left part (of length $x$) to the right with force $T$. Wait, the tension is an internal force. If we consider the section of length $x$ from the end where F is applied, the forces on this section are F (applied externally) and the tension T at the cut pulling this section towards the remaining part of the rod. The acceleration is to the right. So, the net force on this section is $F - T$ (assuming T acts towards the left, opposing F, which is incorrect for a pulling force). Let's re-examine the FBD in the image. The image shows F acting to the right, and T acting to the left on the section of length x. This implies T is the force exerted by the right part of the rod on the left part (length x). If the rod is being pulled by F, the tension T at distance x is the force required to pull the remaining part of the rod (length L-x). Let's consider the section of length L-x from the other end. The only force on this section is the tension T at distance x. The mass of this section is $m_{L-x} = (M/L)(L-x)$. Applying Newton's law to this section: $T = m_{L-x} a = (M/L)(L-x) a$. Substituting $a = F/M$, we get $T = (M/L)(L-x)(F/M) = (L-x)/L * F = (1 - x/L)F$. This is the tension at distance x from the end opposite to where F is applied.

   Let's follow the provided solution's FBD and equation. The provided FBD shows force F acting right and tension T acting left on the section of length x from the end where F is applied. The equation is $F - T = (\frac{M}{L})x \cdot a$. This implies that the net force $F-T$ causes the acceleration $a$ of the mass $(\frac{M}{L})x$. This setup is correct if T is the force exerted by the part of the rod to the right of the cut on the part of the rod to the left of the cut (length x). Since the part of length x is accelerating to the right, the force F must be greater than the force T.

3. Applying Newton's second law to the section:
   $F - T = m_x a$
   $F - T = \left(\frac{M}{L}\right)x \cdot a$

4. Substituting the acceleration:
   Substitute $a = \frac{F}{M}$ into the equation:
   $F - T = \left(\frac{M}{L}\right)x \left(\frac{F}{M}\right)$
   $F - T = \frac{x}{L}F$

5. Solving for tension T:
   $T = F - \frac{x}{L}F$
   $T = F\left(1 - \frac{x}{L}\right)$

This formula gives the tension in the rod at a distance $x$ from the end where the force $F$ is applied.

Explanation of the solution:

1. Calculate the acceleration of the entire rod using $a = F/M$.
2. Consider the section of the rod of length $x$ from the end where force $F$ is applied. Its mass is $(M/L)x$.
3. Apply Newton's second law to this section. The net force on this section is $F$ (applied) minus the tension $T$ (exerted by the rest of the rod). So, $F - T = (M/L)x \cdot a$.
4. Substitute the value of $a$ and solve for $T$.

Answer: The tension in the rod at a distance 'x' from the end where force is applied is $T = F\left(1 - \frac{x}{L}\right)$.