A 10 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kg package on the ground. Part a) What is the magnitude of the least acceleration the monkey must have if it is to lift the package off the ground.
Professor's solution:
The force the monkey pulls downward on the rope has magnitude F. According to
Newton’s third law, the rope pulls upward on the monkey with a force of the same
magnitude, so Newton’s second law for forces acting on the monkey leads to eq(1)
$$F-m_mg=m_ma_m$$
where $m_m$ is the mass of the monkey and $a_m$ is its acceleration. Since the rope is massless
$F = T$ is the tension in the rope. The rope pulls upward on the package with a force of
magnitude F, so Newton’s second law for the package is
$$F+F_N -m_g=m_pa_p$$
where $m_p$ is the mass of the package, $a_p$ is its acceleration, and $F_N$ is the normal force exerted by the ground on it. Now, if F is the minimum force required to lift the package,
then
$$ F_N = 0~\text{ and }~a_p = 0$$
According to the second law equation for the package, this means
$$F = m_pg$$ Substituting $m_pg$ for $F$ in the equation for the monkey, we solve for $a_m$:
$$a=\frac{F-m_mg}{m_m}=\frac{(m_p-m_m)g}{m_m} = 4.9 m/s^2$$
Two questions on this:
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Why in the solution given above in eq(1) $F$ has a positive sign and the $m_mg$ has a negative one, shouldn't it be the other way round since the motion is anticlockwise so we take weight to be in the direction of motion this holding a positive sign and force a negative sign?
-
The other thing is it really that $T=F$, I thought that to find the acceleration, we shouldn't worry about tension because it will cancel anyway? Or is the F here a pulling force like any pulling force in normal life?
One more thing why did he consider $a_p$ zero?
Best Answer
Answer to 1.: When the motion occurs one-dimensionally, you can assign the direction by giving '+' sign to a part of the line, while assigning '-' to the other. It depends upon you what you choose. Here, the upward direction is taken as '+'ve while the downward direction is taken as '-'ve. And so the weight of the monkey is negative here. But if you chose the downward direction as '+'ve, the upward direction would be '-'ve. All the signs would change then. But it will not bother any physics. But be sure, you assign the correct sign to the forces after deciding the positive and negative direction.
Answer to 2.: No rope in the real world is massless. But in theory, it can be taken as negligible so that tension is same everywhere. But if the string is having considerable mass, tension in it is different at different points.
Reply to your query:
To clarify your confusion, let us take a small example. Suppose, like the professor , we take downward direction as negative. Therefore the acceleration due to gravity will be taken as negative. Now, if we take the downward direction as positive, the acceleration of gravity of the same body will be taken as positive. Magnitude will remain same. Signs are just telling which direction the force is acting.
Now, for your second query, let us take an example of lifting a body against gravity. Let you lift up a tomato of mass $m$ to height $h$ from the ground. Now you apply a force $F_{app}$ which is greater than the gravitational force of the earth on tomato ie. $F_g$ . $F_g$ will do work against the gravity to lift the tomato. The remaining force $F_{app} - F_g$ will accelerate the tomato by increasing its kinetic energy. But if you want minimum effort to lift up the tomato, you will only give that force required only to work against the gravity to lift the tomato up; then it will go up with constant velocity(at the beginning you have to apply force infinitesimally greater than $F_g$ to atleast move the tomato from rest to a certain velocity. Then to lift it in minimum effort, you only will have to apply force equal to $F_g$) . Coming to your case, the extra force would create acceleration $a_p$ . But,like in the above case, in order to do the job with minimum force, you will only have to apply the force equal to the opposing force and in that case, the body will move with constant velocity. As there is no extra part of your force for the acceleration, $a_p$ is thus taken as $0$ .
Hope it helps!