Question

Question: In Figure P5.101, the incline has mass M and is fastened to the stationary horizontal tabletop. The block of mass m is placed near the bottom of the incline and is released with a quick push that sets it sliding upward. The block stops near the top of the incline as shown in the figure and then slides down again, always without friction. Find the force that the table top exerts on the incline throughout this motion in terms of m, M, gand θ.

Short answer

The force exerted by the table top on the incline throughout the motion is

$mg\cos \theta \sin \theta \widehat{i}+\left(Mg+mg{\cos }^2\theta \right)\widehat{j}$

Step-by-step solution

Writing the given data from the question

The mass of incline is M.

The mass of the block is m.

The height of the table is H.

The height of incline is h.

Determining the required formulae

Newton's second law of motion states that acceleration occurs when a force acts on an object.

The expression to calculate the force on the object is given as follows.

$\mathit{F}\mathbf{=}\mathit{m}\mathit{a}$

Here, F is the working acting on object, m is the mass of the object, and a is the acceleration of the object.

Calculating the force exerted by the table top on the incline throughout the motion

Consider the free body diagram as shown below.

Calculate the normal force on the block.

$n=mg\cos \theta$

Consider the forces along the x - axis.

$\begin{array}{l}\sum F_x=m{a}_x\\ R_x-n\sin \theta =0\\ R_x=n\sin \theta \end{array}$

Substitute $mg\cos \theta$ for n into above equation.

$\begin{array}{c}{R}_x=\left(mg\cos \theta \right)\sin \theta \\ =mg\cos \theta \sin \theta \end{array}$

Consider the forces along the y - axis.

$\begin{array}{l}\sum F_y=m{a}_y\\ -Mg-n\cos \theta +R_y=0\\ R_y=Mg+n\cos \theta \end{array}$

Substitute $mg\cos \theta$ for n into above equation.

$\begin{array}{c}{R}_y=Mg+\left(mg\cos \theta \right)\cos \theta \\ =Mg+mg{\cos }^2\theta \end{array}$

Calculate the force exerted on the table top.

$\overrightarrow{R}=R_x\widehat{i}+R_y\widehat{j}$

Substitute $mg\cos \theta \sin \theta$ for R_x and $Mg+mg{\cos }^2\theta$ for R_y into above equation.

$\overrightarrow{R}=mg\cos \theta \sin \theta \widehat{i}+\left(Mg+mg{\cos }^2\theta \right)\widehat{j}$

Hence, the force exerted by the table top on the incline throughout the motion is .

$mg\cos \theta \sin \theta \widehat{i}+\left(Mg+mg{\cos }^2\theta \right)\widehat{j}$