Question

Three blocks with masses m1 = 4.08kg, m2 = 2.07kg, and m3 = 17.3kg are shown in the diagram. What must the value of |F| be to keep m1 and m2 stationary with respect to m3?

Answer 1

there are two ways of solving this problem the easiest method is D'Alembert's principle. But let me do it first in newton's second law of motion,

First we must find the acceleration of the m3 let's assume it as a1,
then to find that we apply,

$$\begin{align*}\rightarrow F&=ma\\
\end{align*}$$
to m2
$$R=m_3a$$to m3,
$$\begin{align*}F-R&=m_3a\\
F-m_2a&=m_3a\\
a&=\frac{F}{m_3+m_2}\end{align*}$$
for m1 to move with the m3, m1 and m3 should have the same acceleration thus,
to m1
$$\rightarrow F=ma\\
T=m_1a$$Since m2 should be stationary in vertically,
to m2
$$\downarrow F=ma\\
m_2g-T=m_2(0)\\
T=m_2g$$
this gives,
$$m_2g=m_1a\implies a=\frac{m_2}{m_1}g$$
Now we have two expressions for a, by equating them together,
$$\frac{m_2}{m_1}g=\frac{F}{m_3+m_2}\implies F=\frac{(m_3+m_2)m_2}{m_1}g$$

Answer 2

If you haven't done the D'alembert principle I recommend that you should ignore this, Becuase, if you do, you might mix them up.

The other method is the use of D'alembert principle. We apply each force of $$m_ia$$ where m=1,2,3 for each object as shown in the diagram. Now according to the principle the objects are stationary(or static). thus
$$\sum \vec F=0$$ for any object.
(please consider that only a-acceleration disappears in the system and all the other forces still exists)
For m_1,
$$\rightarrow T-m_1a=0\implies T=m_1a$$For m_2,
$$\downarrow m_2g-T=0\implies m_2g=T=m_1a\implies a=\frac{m_2}{m_1}g$$
for m_2,
$$\rightarrow R-m_2a=0 \implies R=m_2a$$for m_3,
$$\rightarrow F-m_3a-R=0 \implies F=m_3a+(m_2a)=a(m_3+m_2)$$
$$F=\frac{m_2(m_3+m_2)}{m_1}g$$

Answer 3

Not the correct answer, but we'll take it.