The coefficients of friction between blocks A and C and the horizontal surfaces are μs = 0.24 and μk = 0.20. Knowing that mA = 5 kg, mB = 10 kg, and mC = 10 kg, determine (a) the tension in the cord, (b) the acceleration of each block.

Question

The coefficients of friction between blocks A and C and the horizontal surfaces are [latex]{ \mu }_{ s }=0.24[/latex] and [latex]{ \mu }_{ k }=0.20[/latex]. Knowing that [latex]{ m }_{ A }=5 \ kg,{ m }_{ B }=10 \ kg,[/latex] and [latex] { m }_{ C }=10 \ kg[/latex], determine (a) the tension in the cord, (b) the acceleration of each block.

Accepted Answer

We first check that static equilibrium is not maintained:

[latex]\begin{aligned}\left(F_{A} ight)_{m}+\left(F_{C} ight)_{m} & =\mu_{s}\left(m_{A}+m_{C} ight) g \& =0.24(5+10) g \& =3.6 g\end{aligned}[/latex]

Since [latex]W_{B}=m_{B} g=10 \mathrm{~g}>3.6 \mathrm{~g}[/latex], equilibrium is not maintained.

Block A: [latex] \quad \Sigma F_{y}: \quad N_{A}=m_{A} g[/latex]

[latex]F_{A}=\mu_{k} N_{A}=0.2 m_{A} g[/latex]

[latex]\overset{+}{\longrightarrow } \Sigma F_{\lambda}=m_{A} a_{A}: \quad T-0.2 m_{A} g=m_{A} a_{A}[/latex] (1)

Block C: [latex]\quad \Sigma F_{y}: \quad N_{C}=m_{C} g[/latex]

[latex]F_{C}=\mu_{k} N_{C}=0.2 m_{C} g[/latex]

[latex]\overset{+}{\longleftarrow } \Sigma F_{x}=m_{C} a_{C}: T-0.2 m_{C} g=m_{C} a_{C}[/latex] (2)

Block B: [latex]\quad+\downarrow \Sigma F_{y}=m_{B} a_{B}[/latex]

[latex]m_{B} g-2 T=m_{B} a_{B}[/latex] (3)

From kinematics: [latex]\quad a_{B}=\frac{1}{2}\left(a_{A}+a_{C} ight)[/latex] (4)

(a) Tension in cord. Given data: [latex]m_{A}=5 \mathrm{~kg}[/latex]

[latex]m_{B}=m_{C}=10 \mathrm{~kg}[/latex]

Eq. (1): [latex]T-0.2(5) g=5 a_{A} \quad a_{A}=0.2 T-0.2 g[/latex] (5)

Eq. (2): [latex]T-0.2(10) g=10 a_{C} \quad a_{C}=0.1 T-0.2 g[/latex] (6)

Eq. (3): [latex]10 g-2 T=10 a_{B} \quad a_{B}=g-0.2 T[/latex] (7)

Substitute into (4):

[latex]\begin{aligned}g-0.2 T & =\frac{1}{2}(0.2 T-0.2 g+0.1 T-0.2 g) \1.2 g & =0.35 T \quad T=\frac{24}{7} g=\frac{24}{7}\left(9.81 \mathrm{~m} / \mathrm{s}^{2} ight)\end{aligned}[/latex]

[latex]T=33.6 \mathrm{~N}\blacktriangleleft[/latex]

(b) Substitute for T into (5), (7), and (6):

[latex]\begin{array}{lr}a_{A}=0.2\left\lgroup\frac{24}{7} g ight group-0.2 g=0.4857\left(9.81 \mathrm{~m} / \mathrm{s}^{2} ight) & \mathbf{a}_{A}=4.76 \mathrm{~m} / \mathrm{s}^{2} \longrightarrow\blacktriangleleft \a_{B}=g-0.2\left\lgroup\frac{24}{7} g ight group=0.3143\left(9.81 \mathrm{~m} / \mathrm{s}^{2} ight) & \mathbf{a}_{B}=3.08 \mathrm{~m} / \mathrm{s}^{2}\downarrow \blacktriangleleft \a_{C}=0.1\left\lgroup\frac{24}{7} g ight group-0.2 g=0.14286\left(9.81 \mathrm{~m} / \mathrm{s}^{2} ight) & \mathbf{a}_{C}=1.401 \mathrm{~m} / \mathrm{s}^{2}\longleftarrow \blacktriangleleft\end{array}[/latex]

Question 12.30: The coefficients of friction between blocks A and C and the ...

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