Two blocks made of different materials, connected by a thin cord, slide down a plane ramp inclined at an angle$\theta$to the horizontal, Fig. 4–76 (block B is above block A). The masses of the blocks are$m_{\mathrm{A}}$and$m_{\mathrm{B}}$, and the coefficients of friction are${\mu }_{\mathrm{A}}$and${\mu }_{\mathrm{B}}$. If$m_{\mathrm{A}}=m_{\mathrm{B}}=5\mathrm{kg}$and${\mu }_{\mathrm{A}}=0.20$and${\mu }_{\mathrm{B}}=0.30$, determine (a) the acceleration of the blocks and (b) the tension in the cord, for an angle$\theta =32°$.

The coefficient of friction of the upper block is more. The block will drag behind the lower block, and as a result, tension force will develop.

Also, both of the blocks will have identical acceleration.

Given data:

The mass of the snowboarder is $m_{\mathrm{A}}=m_{\mathrm{B}}=5\mathrm{kg}$.

The coefficient of friction for block A is ${\mu }_{\mathrm{A}}=0.20$.

The coefficient of friction for block B is ${\mu }_{\mathrm{B}}=0.30$.

The angle is $\theta =32°$.

The free body diagram of the blocks is as follows:

Consider the x-direction to be along the plane, and the y-direction to be perpendicular to the plane.

The relation to calculate the forces in the vertical direction for block A is given by:

$\begin{array}{c}\Sigma F_{\mathrm{y}}=0\\ F_{\mathrm{NA}}-m_{\mathrm{A}}g\cos \theta =0\\ F_{\mathrm{NA}}=m_{\mathrm{A}}g\cos \theta \end{array}$

Here, $F_{\mathrm{NA}}$ is the normal force for block A and g is the gravitational acceleration.

The relation of force in the horizontal direction for block A is given by:

role="math" localid="1645769479745" $\begin{array}{c}\Sigma F_{\mathrm{x}}=m_{\mathrm{A}}a\\ m_{\mathrm{A}}g\sin \theta -F_{\mathrm{frA}}-F_{\mathrm{T}}=m_{\mathrm{A}}a\\ m_{\mathrm{A}}a=m_{\mathrm{A}}g\sin \theta -{\mu }_{\mathrm{A}}{F}_{\mathrm{NA}}-F_{\mathrm{T}}\\ m_{\mathrm{A}}a=m_{\mathrm{A}}g\sin \theta -{\mu }_{\mathrm{A}}{m}_{\mathrm{A}}g\cos \theta -F_{\mathrm{T}}\dots \left(i\right)\end{array}$

Here, $F_{\mathrm{T}}$is the tension force and $F_{\mathrm{frA}}$is the frictional force for block A.

The relation to calculate the forces in the vertical direction for block B is given by:

$\begin{array}{c}\Sigma {F\text{'}}_y=0\\ F_{\mathrm{NB}}-m_{\mathrm{B}}g\cos \theta =0\\ F_{\mathrm{NB}}=m_{\mathrm{B}}g\cos \theta \end{array}$

Here, $F_{\mathrm{NB}}$ is the normal force for block B.

The relation of force in the horizontal direction for block B is given by:

$\begin{array}{c}\Sigma {F\text{'}}_{\mathrm{x}}=m_{\mathrm{B}}a\\ m_{\mathrm{B}}g\sin \theta -F_{\mathrm{frB}}+F_{\mathrm{T}}=m_{\mathrm{B}}a\\ m_{\mathrm{B}}a=m_{\mathrm{B}}g\sin \theta -{\mu }_{\mathrm{B}}{F}_{\mathrm{NB}}+F_{\mathrm{T}}\end{array}$

$m_{\mathrm{B}}a=m_{\mathrm{B}}g\sin \theta -{\mu }_{\mathrm{B}}{m}_{\mathrm{B}}g\cos \theta +F_{\mathrm{T}}\dots \left(ii\right)$

Here, $F_{\mathrm{frB}}$is the frictional force for block B.

On adding equations (i) and (ii), you get:

$\begin{array}{c}{m}_{\mathrm{A}}a+m_{\mathrm{B}}a=\left(m_{\mathrm{A}}g\sin \theta -{\mu }_{\mathrm{A}}{m}_{\mathrm{A}}g\cos \theta -F_{\mathrm{T}}\right)+\left(m_{\mathrm{B}}g\sin \theta -{\mu }_{\mathrm{B}}{m}_{\mathrm{B}}g\cos \theta +F_{\mathrm{T}}\right)\\ a=\frac{\left[m_{\mathrm{A}}\left(\sin \theta -{\mu }_{\mathrm{A}}\cos \theta \right)+m_{\mathrm{B}}\left(\sin \theta -{\mu }_{\mathrm{B}}\cos \theta \right)\right]g}{\left(m_{\mathrm{A}}+m_{\mathrm{B}}\right)}\\ a=\left[\frac{\left[\begin{array}{l}\left(5\mathrm{kg}\right)\left(\sin 32°-\left(0.20\right)\cos 32°\right)+\\ \left(5\mathrm{kg}\right)\left(\sin 32°-\left(0.30\right)\cos 32°\right)\end{array}\right]\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)}{\left(5\mathrm{kg}+5\mathrm{kg}\right)}\right]\\ a=3.1\mathrm{m}/{\mathrm{s}}^2\end{array}$

Thus, $a=3.1\mathrm{m}/{\mathrm{s}}^2$is the acceleration of the block.

On using equation (i) to calculate the tension force, you get:

$\begin{array}{c}{m}_{\mathrm{A}}a=m_{\mathrm{A}}g\sin \theta -{\mu }_{\mathrm{A}}{m}_{\mathrm{A}}g\cos \theta -F_{\mathrm{T}}\\ F_{\mathrm{T}}=m_{\mathrm{A}}\left(g\sin \theta -{\mu }_{\mathrm{A}}g\cos \theta -a\right)\end{array}$

Substitute the known values in the above relation.

$\begin{array}{c}{F}_{\mathrm{T}}=\left(5\mathrm{kg}\right)\left[\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\sin 32°-\left(0.20\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\cos 32°-3.1\mathrm{m}/{\mathrm{s}}^2\right]\\ F_{\mathrm{T}}=2.15\mathrm{N}\end{array}$

Thus, $F_{\mathrm{T}}=2.15\mathrm{N}$is the required tension force.