Q113CP - University Physics with Modern Physics

Question

A wedge with mass M rests on a frictionless, horizontal table top. A block with mass   is placed on the wedge, and a horizontal force $\overrightarrow{\mathbf{F}}$ is applied to the wedge (Fig. P5.112b). What must the magnitude of   be if the block is to remain at a constant height above the tabletop?

Step-by-Step Solution

Answer

The required magnitude of the force is $\mathrm{F}=(\mathrm{M}+\mathrm{m})\mathrm{g} \tan \mathrm{\alpha }$.

Step 1: identification of given data.

• The mass of the wedge is M.
• The mass of the block is m.

Step 2: Concept/Significance of Newton’s second law:

From the Newton's second law of motion, the net force $\mathbf{\left(}{\mathbf{F}}_{\mathbf{net}}\mathbf{\right)}$ acting on a block is equal to the mass (m) times the acceleration (a) of the block.

${\mathbf{F}}_{\mathbf{net}}\mathbf{=}\mathbf{ma}$ 

Step 3: Find the magnitude of F → :

Draw the free-body diagram of wedge and block.

 

A wedge with mass M rests on a frictionless horizontal table top. A block with mass m is placed on the wedge. The masses M  and m are moving with same acceleration a  and the direction of all forces on masses are shown below figure.

 

Let F be the magnitude of horizontal force acting on the wedge along horizontal direction.

 

From the above figure, the net horizontal force acting on the block is given by,

$(F_{net}{)}_{\mathrm{x}}=\mathrm{Nsin} \mathrm{\alpha }$                                                                                                    ….. (1)

Here, N  is normal force exerted by wedge on the block, and $\alpha$ be the angle of inclination.

 

 

From the Newton's second law of motion, the horizontal force acting on the block is given by,

$(F_{net}{)}_x=\mathrm{ma}$ 

 

Substitute ma  for ${\left({\mathrm{F}}_{\mathrm{net}}\right)}_{\mathrm{x}}$ into equation (1).

$ma=N \sin \alpha$                                                                                                         ….. (2)

 

The net vertical force acting on the block due to vertical upward force    and vertical downward weight mg is given by,

$(F_{net}{)}_y=Ncos \alpha -mg$                                                                                             ….. (3)

 

For no motion along vertical the net vertical force acting on the block should be zero.

$(F_{net}{)}_y=0$ 

 

Substitute 0 for ${\left(F_{net}\right)}_y$ in equation (3).

$0=N \cos \alpha -mg$ 

$mg=N \cos \alpha$                                                                                                        ….. (4)

 

Divide equation (2) with equation (4).

$$\begin{gathered} \frac{mg}{mg}=\frac{N \sin \alpha }{N \cos \alpha } \\ a=g \tan \alpha \end{gathered}$$ 

 

From Newton’s second law of motion, the net external force acting on the total mass of block and wedge is given by,

$F=(M+m)a$ 

Substitute $g \tan \alpha$ for a in the above equation.

$F=(M+m)g \tan \alpha$ 

 

Therefore, the required magnitude of the force is $F=(M+m)g \tan \alpha$.