Question30. (III) For two blocks, connected by a cord and sliding down the incline shown in Fig. 5-40 (see Problem 29), describe the motion $(a)$ if $\mu_{\mathrm{A}}<\mu_{\mathrm{B}}$ , and $(b)$ if $\mu_{\mathrm{A}}>\mu_{\mathrm{B}}$ . (c) Determine a formula for the acceleration of each block and the tension $F_{\mathrm{T}}$ in the cord in terms of $m_{\mathrm{A}}, m_{\mathrm{B}}$ , and $\theta$ ; interpret your results in light of your answers to $(a)$ and $(b)$

Studdy Solution

STEP 1

1. Two blocks $A$ and $B$ are connected by a cord and are sliding down an incline.
2. The incline makes an angle $\theta$ with the horizontal.
3. The coefficients of friction for blocks $A$ and $B$ are $\mu_A$ and $\mu_B$ respectively.
4. We need to analyze the motion of the blocks based on the relative values of $\mu_A$ and $\mu_B$ .
5. We need to determine the formula for acceleration and tension in the cord.

STEP 2

1. Analyze the motion when $\mu_A < \mu_B$ .
2. Analyze the motion when $\mu_A > \mu_B$ .
3. Derive the formula for acceleration and tension in the cord.
4. Interpret the results in light of the analyses in steps 1 and 2.

STEP 3

When $\mu_A < \mu_B$ , block $A$ experiences less friction compared to block $B$ . This means block $A$ will tend to slide down the incline faster than block $B$ . As a result, block $B$ will exert a tension force on block $A$ through the cord, attempting to slow down block $A$ and accelerate block $B$ .

STEP 4

When $\mu_A > \mu_B$ , block $B$ experiences less friction compared to block $A$ . This means block $B$ will tend to slide down the incline faster than block $A$ . As a result, block $A$ will exert a tension force on block $B$ through the cord, attempting to slow down block $B$ and accelerate block $A$ .

STEP 5

To derive the formula for acceleration and tension, consider the forces acting on each block. For block $A$ , the net force along the incline is: $m_A g \sin \theta - \mu_A m_A g \cos \theta - T = m_A a$
For block $B$ , the net force along the incline is: $m_B g \sin \theta - \mu_B m_B g \cos \theta + T = m_B a$
Solving these two equations simultaneously will give the expressions for acceleration $a$ and tension $T$ .

STEP 6

Add the two equations to eliminate $T$ : $(m_A + m_B) g \sin \theta - (\mu_A m_A + \mu_B m_B) g \cos \theta = (m_A + m_B) a$
Solve for $a$ : $a = \frac{(m_A + m_B) g \sin \theta - (\mu_A m_A + \mu_B m_B) g \cos \theta}{m_A + m_B}$
Substitute $a$ back into one of the original equations to solve for $T$ : $T = m_A g \sin \theta - \mu_A m_A g \cos \theta - m_A a$

STEP 7

Interpretation: - If $\mu_A < \mu_B$ , the acceleration $a$ will be positive, indicating that the system accelerates down the incline. - If $\mu_A > \mu_B$ , the acceleration $a$ will still be positive, but the tension $T$ will be different, reflecting the different frictional forces.
The formulas for acceleration and tension are: $a = \frac{(m_A + m_B) g \sin \theta - (\mu_A m_A + \mu_B m_B) g \cos \theta}{m_A + m_B}$ $T = m_A g \sin \theta - \mu_A m_A g \cos \theta - m_A a$

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