GREPhysics.NET: GR0177 #73

GR0177 #73

Problem

GREPhysics.NET Official Solution

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Mechanics $\Rightarrow$ }Forces

The net force of the system is $\sum F = (m_a+m_b) a$ , and thus the net acceleration due to this force is $a = F/(m_a+m_b)$ .

The net force for mass A is just $m_a a = F - F_{ab}$ . By Newton's Third Law, $F_{ab}=N$ is just the normal force exerted by A on B.

Solving for the normal force, one finds that $N = F - m_a a = F(1 - m_a/(m_a+m_b))=F(m_b/(m_a+m_b))$ .

Summing up the vertical forces on mass B, and using the fact that the frictional force is just $f=\mu N$ , one finds that $\sum F_y = \mu N-m_b g=0$ for the applied force to balance its mass completely. Thus, $F= (m_a+m_b)g/\mu$ , which is approximately 400N, as in choice (D) after plugging in the numbers.

Alternate Solutions

R0BB23
2008-06-30 08:46:19

$\sum F_y = \mu N-m_b g=0$
$\mu N=m_b g$
$N=m_b a$
$a=g/\mu$
$F=(m_a+m_b)a$
$F=(m_a+m_b)g/\mu$

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Comments

carl_the_sagan 2008-11-07 22:05:17	I'm not sure if this is a valid way to tackle the problem but it's how my messed up line of thinking went: The block B needs to be pushed up equal to the force gravity exerts on it to stay in place. Take gravity to be 10m/s^2 for convenience and it needs 40N. However, the coefficient of friction is only .5, so you have to push twice as hard, so 80N. This means you need to accelerate B by 80N/4kg = 20m/s^2 horizontally to produce enough "lift" to balance things. Since the entire system is 20kg, F = 20kg*20m/s^2 = 400N Reply to this comment
R0BB23 2008-06-30 08:46:19	$\sum F_y = \mu N-m_b g=0$ $\mu N=m_b g$ $N=m_b a$ $a=g/\mu$ $F=(m_a+m_b)a$ $F=(m_a+m_b)g/\mu$ Reply to this comment

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$\sum F_y = \mu N-m_b g=0$ $\mu N=m_b g$ $N=m_b a$ $a=g/\mu$ $F=(m_a+m_b)a$ $F=(m_a+m_b)g/\mu$

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