- #1
PhyAmateur
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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 says: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$$ F_N = 0 $$and$$a_p = 0$$. According to the second law equation for the package, this means
$$F = m_pg$$ Substituting mpg for F in the equation for the monkey, we solve for $$a_m$$:
$$a=\frac{F-m_mg}{m_m}=\frac{(m_p-m_m)g}{m_m} = 4.9 m/s^2$$
2 questions on this: Why in the solution given above in eq(1) the 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 $$a_p$$ negative?
Professor's solution says: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$$ F_N = 0 $$and$$a_p = 0$$. According to the second law equation for the package, this means
$$F = m_pg$$ Substituting mpg for F in the equation for the monkey, we solve for $$a_m$$:
$$a=\frac{F-m_mg}{m_m}=\frac{(m_p-m_m)g}{m_m} = 4.9 m/s^2$$
2 questions on this: Why in the solution given above in eq(1) the 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 $$a_p$$ negative?