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By Newton's third law, if the rope is pulling on the boat with some force, then the boat is pulling on the rope with this same force. So either terminology is fine. By attaching the rope to the boat and by there being tension in the rope, the rope and boat are now pulling on each other.

Your proposition that $T\leq T_0e^{-\mu\theta}$ doesn't make sense, as it suggests that the more the rope is wrapped around the pole the smaller the tension can be before slipping.

To see why $T(\theta +\text d\theta)\leq T(\theta)+\mu T\text d\theta$ is correct, let's look at a simpler scenario with static friction. Let's say I have a block on a flat surface with friction, and one force $T_1$ is pulling on the block to the left, and another force $T_2$ is pulling on the block to the right. If $T_2\neq T_1$ but the block is not moving, it must be that $$|T_2-T_1|\leq\mu N$$ However, if we know what the direction of impending motion is, we can get rid of the absolute value sign. For example, if we know friction preventing the block from sliding to the right, then we know that $T_2>T_1$, and so we have $$T_2-T_1\leq \mu N$$

The same thing is happening here. We are assuming $\text dT=T(\theta+\text d\theta)-T(\theta)>0$ so that impending motion is in the direction of $\text d\theta>0$. This is why we have $T(\theta +\text d\theta)\leq T(\theta)+\mu T\text d\theta$.

Note that none of this work determines actual tensions in the system. All this work shows is the limit to $T$ before slipping occurs given values for $T_0$, $\mu$, and $\theta$.

As for why $\text dT>0$, you are right that this isn't always true, just like how in my example it doesn't have to be the case that $T_2>T_1$. To set the sign of $\text dT$ we need to either assume or reason to the direction of impending motion. The problem obviously assumes impending motion in the direction of increasing $\theta$, which I think is reasonable. I suppose it would have been better for the question to explicitly state this assumption in more detail though.

If this is still unsatisfying, then let's get technical: which is larger for $\theta>0$, $T_0e^{\mu\theta}$, or $T_0e^{-\mu\theta}$?

By Newton's third law, if the rope is pulling on the boat with some force, then the boat is pulling on the rope with this same force. So either terminology is fine. By attaching the rope to the boat and by there being tension in the rope, the rope and boat are now pulling on each other.

Your proposition that $T\leq T_0e^{-\mu\theta}$ doesn't make sense, as it suggests that the more the rope is wrapped around the pole the smaller the tension can be before slipping.

To see why $T(\theta +\text d\theta)\leq T(\theta)+\mu T\text d\theta$ is correct, let's look at a simpler scenario with static friction. Let's say I have a block on a flat surface with friction, and one force $T_1$ is pulling on the block to the left, and another force $T_2$ is pulling on the block to the right. If $T_2\neq T_1$ but the block is not moving, it must be that $$|T_2-T_1|\leq\mu N$$ However, if we know what the direction of impending motion is, we can get rid of the absolute value sign. For example, if we know friction is preventing the block from sliding to the right, then we know that $T_2>T_1$, and so we have $$T_2-T_1\leq \mu N$$

The same thing is happening here. We are assuming $\text dT=T(\theta+\text d\theta)-T(\theta)>0$ so that impending motion is in the direction of $\text d\theta>0$. This is why we have $T(\theta +\text d\theta)\leq T(\theta)+\mu T\text d\theta$.

Note that none of this work determines actual tensions in the system. All this work shows is the limit to $T$ before slipping occurs given values for $T_0$, $\mu$, and $\theta$.

As for why $\text dT>0$, you are right that this isn't always true, just like how in my example it doesn't have to be the case that $T_2>T_1$. To set the sign of $\text dT$ we need to either assume or reason to the direction of impending motion. The problem obviously assumes impending motion in the direction of increasing $\theta$, which I think is reasonable. I suppose it would have been better for the question to explicitly state this assumption in more detail though.

If this is still unsatisfying, then let's get technical: which is larger for $\theta>0$, $T_0e^{\mu\theta}$, or $T_0e^{-\mu\theta}$?

By Newton's third law, if the rope is pulling on the boat with some force, then the boat is pulling on the rope with this same force. So either terminology is fine. By attaching the rope to the boat and by there being tension in the rope, the rope and boat are now pulling on each other.

Your proposition that $T\leq T_0e^{-\mu\theta}$ doesn't make sense, as it suggests that the more the rope is wrapped around the pole the smaller the tension can be before slipping.

To see why $T(\theta +\text d\theta)\leq T(\theta)+\mu T\text d\theta$ is correct, let's look at a simpler scenario with static friction. Let's say I have a block on a flat surface with friction, and one force $T_1$ is pulling on the block to the left, and another force $T_2$ is pulling on the block to the right. If $T_2\neq T_1$ but the block is not moving, it must be that $$|T_2-T_1|\leq\mu N$$ However, if we know what the direction of impending motion is, we can get rid of the absolute value sign. For example, if we know that friction is preventing the block from sliding to the right, then we know that $T_2>T_1$, and so we have $$T_2-T_1\leq \mu N$$

The same thing is happening here. We are assuming $\text dT=T(\theta+\text d\theta)-T(\theta)>0$ so that impending motion is in the direction of $\text d\theta>0$. This is why we have $T(\theta +\text d\theta)\leq T(\theta)+\mu T\text d\theta$.

Note that none of this work determines actual tensions in the system. All this work shows is the limit to $T$ before slipping occurs given values for $T_0$, $\mu$, and $\theta$.

As for why $\text dT>0$, you are right that this isn't always true, just like how in my example it doesn't have to be the case that $T_2>T_1$. To set the sign of $\text dT$ we need to either assume or reason to the direction of impending motion. The problem obviously assumes impending motion in the direction of increasing $\theta$, which I think is reasonable. I suppose it would have been better for the question to explicitly state this assumption in more detail though.

If this is still unsatisfying, then let's get technical: which is larger for $\theta>0$, $T_0e^{\mu\theta}$, or $T_0e^{-\mu\theta}$?

By Newton's third law, if the rope is pulling on the boat with some force, then the boat is pulling on the rope with this same force. So either terminology is fine. By attaching the rope to the boat and by there being tension in the rope, the rope and boat are now pulling on each other.

Your proposition that $T\leq T_0e^{-\mu\theta}$ doesn't make sense, as it suggests that the more the rope is wrapped around the pole the smaller the tension can be before slipping.

To see why $T(\theta +\text d\theta)\leq T(\theta)+\mu T\text d\theta$ is correct, let's look at a simpler scenario with static friction. Let's say I have a block on a flat surface with friction, and one force $T_1$ is pulling on the block to the left, and another force $T_2$ is pulling on the block to the right. If $T_2\neq T_1$ but the block is not moving, it must be that $$|T_2-T_1|\leq\mu N$$ However, if we know what the direction of impending motion is, we can get rid of the absolute value sign. For example, if we know friction preventing the block from sliding to the right, then we know that $T_2>T_1$, and so we have $$T_2-T_1\leq \mu N$$

The same thing is happening here. We are assuming $\text dT=T(\theta+\text d\theta)-T(\theta)>0$ so that impending motion is in the direction of $\text d\theta>0$. This is why we have $T(\theta +\text d\theta)\leq T(\theta)+\mu T\text d\theta$.

Note that none of this work determines actual tensions in the system. All this work shows is the limit to $T$ before slipping occurs given values for $T_0$, $\mu$, and $\theta$.

As for why $\text dT>0$, you are right that this isn't always true, just like how in my example it doesn't have to be the case that $T_2>T_1$. To set the sign of $\text dT$ we need to either assume or reason to the direction of impending motion. The problem obviously assumes impending motion in the direction of increasing $\theta$, which I think is reasonable. I suppose it would have been better for the question to explicitly state this assumption in more detail though.

If this is still unsatisfying, then let's get technical: which is larger for $\theta>0$, $T_0e^{\mu\theta}$, or $T_0e^{-\mu\theta}$?

By Newton's third law, if the rope is pulling on the boat with some force, then the boat is pulling on the rope with this same force. So either terminology is fine. By attaching the rope to the boat and by there being tension in the rope, the rope and boat are now pulling on each other.

Your proposition that $T\leq T_0e^{-\mu\theta}$ doesn't make sense, as it suggests that the more the rope is wrapped around the pole the smaller the tension can be before slipping.

To see why $T(\theta +\text d\theta)\leq T(\theta)+\mu T\text d\theta$ is correct, let's look at a simpler scenario with static friction. Let's say I have a block on a flat surface with friction, and one force $T_1$ is pulling on the block to the left, and another force $T_2$ is pulling on the block to the right. If $T_2\neq T_1$ but the block is not moving, it must be that $$|T_2-T_1|\leq\mu N$$ However, if we know what the direction of impending motion is, we can get rid of the absolute value sign. For example, if we know friction preventing the block from sliding to the right, then we know that $T_2>T_1$, and so we have $$T_2-T_1\leq \mu N$$

The same thing is happening here. We are assuming $\text dT=T(\theta+\text d\theta)-T(\theta)>0$ so that impending motion is in the direction of $\text d\theta>0$. This is why we have $T(\theta +\text d\theta)\leq T(\theta)+\mu T\text d\theta$.

Note that none of this work determines actual tensions in the system. All this work shows is the limit to $T$ before slipping occurs given values for $T_0$, $\mu$, and $\theta$.

As for why $\text dT>0$, you are right that this isn't always true, just like how in my example it doesn't have to be the case that $T_2>T_1$. To set the sign of $\text dT$ we need to either assume or reason to the direction of impending motion. The problem obviously assumes impending motion in the direction of increasing $\theta$, which I think is reasonable. I suppose it would have been better for the question to explicitly state this assumption in more detail though.

If this is still unsatisfying, then let's get technical: which is larger for $\theta>0$, $T_0e^{\mu\theta}$, or $T_0e^{-\mu\theta}$?

By Newton's third law, if the rope is pulling on the boat with some force, then the boat is pulling on the rope with this same force. So either terminology is fine. By attaching the rope to the boat and by there being tension in the rope, the rope and boat are now pulling on each other.

Your proposition that $T\leq T_0e^{-\mu\theta}$ doesn't make sense, as it suggests that the more the rope is wrapped around the pole the smaller the tension can be before slipping.

To see why $T(\theta +\text d\theta)\leq T(\theta)+\mu T\text d\theta$ is correct, let's look at a simpler scenario with static friction. Let's say I have a block on a flat surface with friction, and one force $T_1$ is pulling on the block to the left, and another force $T_2$ is pulling on the block to the right. If $T_2\neq T_1$ but the block is not moving, it must be that $$|T_2-T_1|\leq\mu N$$ However, if we know what the direction of impending motion is, we can get rid of the absolute value sign. For example, if we know that friction is preventing the block from sliding to the right, then we know that $T_2>T_1$, and so we have $$T_2-T_1\leq \mu N$$

The same thing is happening here. We are assuming $\text dT=T(\theta+\text d\theta)-T(\theta)>0$ so that impending motion is in the direction of $\text d\theta>0$. This is why we have $T(\theta +\text d\theta)\leq T(\theta)+\mu T\text d\theta$.

Note that none of this work determines actual tensions in the system. All this work shows is the limit to $T$ before slipping occurs given values for $T_0$, $\mu$, and $\theta$.

As for why $\text dT>0$, you are right that this isn't always true, just like how in my example it doesn't have to be the case that $T_2>T_1$. To set the sign of $\text dT$ we need to either assume or reason to the direction of impending motion. The problem obviously assumes impending motion in the direction of increasing $\theta$, which I think is reasonable. I suppose it would have been better for the question to explicitly state this assumption in more detail though.

If this is still unsatisfying, then let's get technical: which is larger for $\theta>0$, $T_0e^{\mu\theta}$, or $T_0e^{-\mu\theta}$?