Question28. (II) (a) Suppose the coefficient of kinetic friction between $m_{\mathrm{A}}$ and the ramp in Fig. $5-39$ is $\mu_{\mathrm{k}}=0.15$ , and that $m_{\mathrm{A}}=m_{\mathrm{B}}=2.7 \mathrm{~kg}$ . As $m_{\mathrm{B}}$ moves down, determine the magnitude of the acceleration of $m_{\mathrm{A}}$ and $m_{\mathrm{B}}$ , given $\theta=34^{\circ}$ . (b) What smallest value of $\mu_{\mathrm{k}}$ will keep the system from accelerating? [Ignore masses of cord and (frictionless) pulley.]
FIGURE 5-39 Problem 28. Problems 139

Studdy Solution

STEP 1

1. The coefficient of kinetic friction between $m_{\mathrm{A}}$ and the ramp is $\mu_{\mathrm{k}} = 0.15$ .
2. Both masses $m_{\mathrm{A}}$ and $m_{\mathrm{B}}$ are equal to $2.7 \, \text{kg}$ .
3. The angle of the ramp $\theta$ is $34^{\circ}$ .
4. The pulley is frictionless, and the masses of the cord and pulley are ignored.
5. We need to find the acceleration of the system and the smallest value of $\mu_{\mathrm{k}}$ that will keep the system from accelerating.

STEP 2

1. Analyze forces acting on the masses.
2. Write equations of motion for the system.
3. Solve for the acceleration of the system.
4. Determine the smallest value of $\mu_{\mathrm{k}}$ to prevent acceleration.

STEP 3

Analyze forces acting on the masses.
For $m_{\mathrm{A}}$ on the ramp: - Gravitational force component along the ramp: $m_{\mathrm{A}} g \sin \theta$ - Normal force: $N = m_{\mathrm{A}} g \cos \theta$ - Frictional force: $f_{\mathrm{k}} = \mu_{\mathrm{k}} N = \mu_{\mathrm{k}} m_{\mathrm{A}} g \cos \theta$
For $m_{\mathrm{B}}$ : - Gravitational force: $m_{\mathrm{B}} g$

STEP 4

Write equations of motion for the system.
For $m_{\mathrm{A}}$ moving up the ramp: $T - m_{\mathrm{A}} g \sin \theta - f_{\mathrm{k}} = m_{\mathrm{A}} a$
For $m_{\mathrm{B}}$ moving down: $m_{\mathrm{B}} g - T = m_{\mathrm{B}} a$

STEP 5

Solve for the acceleration of the system.
Add the two equations to eliminate $T$ : $m_{\mathrm{B}} g - m_{\mathrm{A}} g \sin \theta - \mu_{\mathrm{k}} m_{\mathrm{A}} g \cos \theta = (m_{\mathrm{A}} + m_{\mathrm{B}}) a$
Solve for $a$ : $a = \frac{m_{\mathrm{B}} g - m_{\mathrm{A}} g \sin \theta - \mu_{\mathrm{k}} m_{\mathrm{A}} g \cos \theta}{m_{\mathrm{A}} + m_{\mathrm{B}}}$
Substitute the given values: $a = \frac{2.7 \times 9.8 - 2.7 \times 9.8 \times \sin 34^{\circ} - 0.15 \times 2.7 \times 9.8 \times \cos 34^{\circ}}{2.7 + 2.7}$
Calculate $a$ .

STEP 6

Determine the smallest value of $\mu_{\mathrm{k}}$ to prevent acceleration.
Set $a = 0$ in the equation: $m_{\mathrm{B}} g = m_{\mathrm{A}} g \sin \theta + \mu_{\mathrm{k}} m_{\mathrm{A}} g \cos \theta$
Solve for $\mu_{\mathrm{k}}$ : $\mu_{\mathrm{k}} = \frac{m_{\mathrm{B}} g - m_{\mathrm{A}} g \sin \theta}{m_{\mathrm{A}} g \cos \theta}$
Substitute the given values: $\mu_{\mathrm{k}} = \frac{2.7 \times 9.8 - 2.7 \times 9.8 \times \sin 34^{\circ}}{2.7 \times 9.8 \times \cos 34^{\circ}}$
Calculate $\mu_{\mathrm{k}}$ .

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