Why does the 1kg mass move towards inclined plane here?

Question

Assumption :All frictionless surfaces.

EDIT: In the book , it doesn’t say friction has to be present but as PeteW has mentioned , i think it should be necessary too. So , in that case let me take two cases. One when friction is present between 5Kg and 1kg block and other when it is not. So , depending upon it. Check it.

The triangle wedge (white color) is always fixed. The blue color wedge of 5kg moves down and so does 1kg is given in Q.

So , the main question to find the time when 1kg block touches the inclined plane. Assume in one case a inertial frame and in another , any non inertial frame of your own choice.

My main confusion is that how do we know if the 1kg block will move towards right direction. It can be left or no direction also.

If I draw the FBD for 1kg block , it has N from 5kg which cancels out due to its mg=1g. So , there is no external force to make it move right.

Answer 1

If you are asking why does the 1 kg mass (I'll denote it as $m_1$), moves towards the inclined plane, then IMHO the exercise you shouldn't have any friction.

• If there is no friction, then the $m_1$ will move downwards, because there will be no horizontal force to keep it attached to the Wedge (I suspect its mass is 5 kg, so I will denote it as $m_5$. Below is the FBD and the kinetic diagrams of the two moving masses $m_1, m_5$.

The acceleration of mass 1 will be downward, while the wedge will accelerate in a direction along the plane.

• If there is friction, then the material will probably never meet the inclined face. The reason is that the friction will be the only force on the horizontal plane, so the $m_1$ will move with $m_5$. (Of course, if you change the slope you might get a reaction so great from the wall, that the friction cannot overcome).

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If you are asking about the forces, then in order to calculate the reactions and the acceleration you'd have to solve the following system

$$\begin{cases} - m_1 \cdot g + N_1 = -m_1\cdot a_1\\ - N_{2x} = - m_5\cdot a_{5x}\\ -m_5 g + N_{2y} - N_1 = - m_5\cdot a_{5y}\\ a_1 = a_{5y}\\ \frac{A_{5x}}{\cos(\phi)} = \frac{a_{5y}}{\sin(\phi)} = a_5\\ \frac{N_{2x}}{\sin(\phi)} = \frac{N_{2y}}{\cos(\phi)} = N_2 \end{cases} $$

If you do the substitutions you obtain the following system with 4 equations and 4 unknowns $a_1, a_5, N_1, N_2 $:

$$\begin{cases} - m_1 \cdot g + \color{red}{N_1} = -m_1\cdot \color{red}{a_1}\\ - \color{red}{N_{2}}\sin(\phi) = - m_5\cdot \color{red}{a_{5}}\cos(\phi)\\ -m_5 g + \color{red}{N_2} \cos(\phi) - N_1= - m_5\cdot\color{red}{ a_{5}}\sin(\phi)\\ \color{red}{a_1} = \color{red}{a_{5}}\sin(\phi)\\ \end{cases} $$

The results (using g= 9.81 $m/s^2$)should be:

• $N_1$: 5.834 N
• $N_2$: 43.82 N,
• $a_1$: 3.975 $m/s^2$
• $a_5$: 6.6 $m/s^2$

I'll leave the numerical solution to any interested party.

Answer 2

Let's assume for now the wedge S massless. its mass is not going the change the aspect of the behavior of the One kg mass. The components of the 1 kg along and perpendicular to the ramp are $$F_{parallel}=mgsin37 =1*9.8*0.601 \quad and \quad F_{perp} =1*9.8cos37$$

The acceleration of the wedge S is

$$\alpha =F_{parallel}/1=gsin37=9.8*0.601=5.89m/s^2$$

And the I kg block will move to the right with the horizontal component of this acceleration because it insulated from the ramp, it just tends to stay in the same x coordinate by its inertia.

And its horizontal acceleration from the frame of reference of the wedge S is,

$$5.89*cos37=4.71m/s^2$$

However, its horizontal acceleration in the frame of reference of the ramp is zero.

Edit

to make an intuitive illustration the 1kg mass will behave as if:

it is dropped into a free fall near a hill with 37 degree slope in a planet with a gravity of 5.89/ 9.8 of the earth. It falls straight down till it crashes on the side of the hill.