Calculating the Horizontal Force for a Non-Slipping Wedge

[ElectricMile]

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




I am a little confused, what is this question really asking? i have attempted sevral trig equations but i can't get the answer correct...anyone point me in the right way?

 

[Doc Al]

First identify all the forces acting on the block. Then apply Newton's 2nd law. (Hint: vertical forces must add to zero, otherwise the block will accelerate in the vertical direction.)

 

[ElectricMile]

im still so confused on this problem...ok i have the forces for the block and the wedge the same, they both have a line pointing up and down, and pointing right for the force...but I am completely lost on how to begin solving this problem.

 

[Leong]

If $$m_1$$ is not to slip in any direction, it will move forward together with $$m_2$$ with accelerelation a. Upward and right direction is choosen to be y and x axis.
Consider $$m_1$$:
y component :
$$N_{1}sin(90-\theta)+(-m_{1}*g)=0$$
$$N_{1}cos\theta=m_{1}*g$$…(1)
x component :
$$N_{1}cos(90-\theta)-m_{1}*a=0$$
$$N_{1}sin\theta=m_{1}*a$$…(2)
Consider $$m_2$$:
x component:
$$F+(-N_{1}sin\theta)=m_{2}*a$$
$$F-N_{1}sin\theta=m_{2}*a$$…(3)
(2) into (3):
$$F-m_{1}*a=m_{2}*a$$
$$F =a*(m_{1}+ m_{2})$$…(4)
(2) divided by (1) :
$$tan\theta=\frac{a}{g}$$…(5)
(5) into (4) :
$$F=gtan\theta(m_{1}+ m_{2})$$