Question

Question: (a) Suppose the coefficient of kinetic friction between $m_A$and the plane in Fig. 4-62 is ${\mu }_k=0.15$and that $m_A=m_{\mathrm{B}}=2.7\mathrm{kg}$. As,$m_B$moves down, determine the magnitude of acceleration of$m_A$and$m_{\mathrm{B}}$, given$\theta ={34}^{\mathrm{o}}$. (b) What smallest value of${\mu }_k$will keep the system from accelerating? [Ignore masses of the (frictionless) pulley and the cord.]

Short answer

(a) The magnitude of the acceleration of the two masses A and B is $1.55\mathrm{m}/{\mathrm{s}}^2$.

(b) The kinetic coefficient of friction for a non-accelerating system is 0.53.

Step-by-step solution

Step 1.Statement of Newton’s second law of motion

Newton’s second law of motion states that the acceleration of a body is directly proportional to the net force acting on the body and inversely proportional to the mass of the body.Mathematically,

$a\propto \frac{F_{\mathrm{net}}}{m}$

Step 2. Drawing the free body diagram of the given situation

A free body diagram depicting all the forces acting on the system is given below.

The forces acting on the two masses are:

• The weight of the mass $m_{\mathrm{A}}$is $m_{\mathrm{A}}g$; downward.
• The weight of the mass $m_{\mathrm{B}}$is $m_{\mathrm{B}}g$; downward.
• The tension in the cord is T.
• The force of friction acting downward along the inclined plane is f.
• The normal force acting on mass $m_{\mathrm{A}}$is N.

Step 3. Identification of given data

The mass of the two blocks is $m_{\mathrm{A}}=m_{\mathrm{B}}=2.7\mathrm{kg}$.

Let their mass be denoted by m.

The coefficient of kinetic friction between the plane and $m_{\mathrm{A}}$is ${\mu }_{\mathrm{k}}=0.15$.

The angle of inclination is $\theta ={34}^{\mathrm{o}}$.

Let the acceleration of the two masses be a.

Step 4. (a) Determination of force equation for block B

From Newton’s law of motion, the equation of force for mass $m_{\mathrm{B}}$is:

role="math" localid="1645601530678" $\begin{array}{c}{m}_{\mathrm{B}}a=\sum F\\ m_{\mathrm{B}}a=m_{\mathrm{B}}g-T\\ T=mg-ma\dots \left(i\right)\end{array}$

Step 5. (a) Determination of force equation for block A

The weight of block A can be resolved into its horizontal and vertical components. From Newton’s second law of motion, the equation of force will be:

$\begin{array}{c}{m}_{\mathrm{A}}a=\sum F\\ m_{\mathrm{A}}a=T-mg\sin \theta -f\\ ma=T-mg\sin \theta -{\mu }_{\mathrm{k}}N\\ ma=T-mg\sin \theta -{\mu }_{\mathrm{k}}\left(mg\cos \theta \right)\dots \left(ii\right)\end{array}$

Substitute equation (i) in equation (ii).

$\begin{array}{c}ma=mg-ma-mg\sin \theta -{\mu }_{\mathrm{k}}\left(mg\cos \theta \right)\\ 2a=g-g\sin \theta -{\mu }_{\mathrm{k}}g\cos \theta \\ a=\frac{g-g\sin \theta -{\mu }_{\mathrm{k}}g\cos \theta }{2}\end{array}$

Substituting the known numerical values in the above expression, you obtain:

$\begin{array}{c}a=\frac{\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)-\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\sin {34}^{\mathrm{o}}-\left(0.15\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\cos {34}^{\mathrm{o}}}{2}\\ =1.55\mathrm{m}/{\mathrm{s}}^2\end{array}$

Thus, the acceleration of the two masses A and B is $1.55\mathrm{m}/{\mathrm{s}}^2$.

Step 6. (b) Determination of the coefficient of friction if the given system is non-accelerating

In a non-accelerating system, the forces opposed to each other should have the same magnitude.

From the FBD, you have:

$\begin{array}{c}mg\sin \theta +{\mu }_{\mathrm{k}}N=mg\\ mg\sin \theta +\left(mg\cos \theta \right){\mu }_{\mathrm{k}}=mg\\ {\mu }_{\mathrm{k}}\cos \theta =1-\sin \theta \\ {\mu }_{\mathrm{k}}=\frac{1-\sin \theta }{\cos \theta }\end{array}$

Substitute the value of the angle of inclination in the above expression.

$\begin{array}{c}{\mu }_{\mathrm{k}}=\frac{1-\sin {34}^{\mathrm{o}}}{\cos {34}^{\mathrm{o}}}\\ =0.53\end{array}$

Thus, the kinetic coefficient of friction for a non-accelerating system is 0.53.