Find an expression for the magnitude of the horizontal force $F$in$FIGUREP7.49$ for which m1 does not slip either up or down along the wedge. All surfaces are frictionless.

the horizontal force acting on the large sliding wedge block is$\stackrel{\rightharpoonup }{F}$

mass of the small sliding block.localid="1649764530848" $m_1$

mass of the large sliding wedge block.$m_2$

the inclination angle of wedge block is $\theta$

All surfaces are frictionless.

Resolving horizontal and vertical forces for the smaller block:

Resolving horizontal forces $\sum _i{F}_{i,x}=0$

role="math" localid="1649765390956" $$\begin{gathered} F_{1,x}-W_{1,x}=0 \\ {m}_{1}a\cos \left(\theta \right)=m_{1}g\sin \left(\theta \right) \\ a=g\tan \left(\theta \right)\left(I\right) \end{gathered}$$

The above equation $\left(I\right)$gives the expression for acceleration of mass $m_1$.

Similarly, resolving vertical forces $\sum _i{F}_{i,y}=0$to get the normal force acting on the smaller block.

$$\begin{gathered} F_N\cos \left(\theta \right)-m_{1}g=0 \\ {F}_N=\frac{m_{1}g}{\cos \left(\theta \right)}\left(II\right) \end{gathered}$$

Now, find the horizontal forces for the system,

$\stackrel{\rightharpoonup }{F}=(m_1+m_2)a$

Substituting the equation $\left(I\right)$in the above expression, we get

$\stackrel{\rightharpoonup }{F}=(m_1+m_2)g\tan \left(\theta \right)$