Math  /  Algebra

Question55. (II) Figure 4-49 shows a block (mass mAm_{\mathrm{A}} ) on a smooth horizontal surface, connected by a thin cord that passes over a pulley to a second block ( mBm_{\mathrm{B}} ), which hangs vertically. (a) Draw a free-body diagram for each block, showing the force of gravity on each, the force (tension) exerted by the cord, and any normal force. (b) Apply Newton's second law to find formulas for the acceleration of the system and for the tension in the cord. Ignore friction and the masses of the pulley and cord.
FIGURE 4-49 Problems 55, 56, and 57. Mass mAm_{\mathrm{A}} rests on a smooth horizontal surface; mBm_{\mathrm{B}} hangs vertically.

Studdy Solution

STEP 1

1. The surface on which block mA m_A rests is smooth, implying no friction.
2. The pulley and cord are massless.
3. The only forces acting on the blocks are gravity, tension, and normal force.
4. We need to find the acceleration of the system and the tension in the cord.

STEP 2

1. Draw free-body diagrams for each block.
2. Apply Newton's second law to each block.
3. Solve the system of equations for acceleration and tension.

STEP 3

Draw free-body diagrams for each block.
- For block mA m_A on the horizontal surface: - Gravity force FgA=mAg F_{gA} = m_A \cdot g acting downward. - Normal force NA N_A acting upward. - Tension force T T acting horizontally to the right.
- For block mB m_B hanging vertically: - Gravity force FgB=mBg F_{gB} = m_B \cdot g acting downward. - Tension force T T acting upward.

STEP 4

Apply Newton's second law to each block.
- For block mA m_A : $ \sum F_x = m_A \cdot a = T \]
- For block mB m_B : $ \sum F_y = m_B \cdot a = m_B \cdot g - T \]

STEP 5

Solve the system of equations for acceleration and tension.
From block mA m_A : T=mAa T = m_A \cdot a
From block mB m_B : mBa=mBgT m_B \cdot a = m_B \cdot g - T
Substitute T=mAa T = m_A \cdot a into the equation for block mB m_B : mBa=mBgmAa m_B \cdot a = m_B \cdot g - m_A \cdot a
Combine terms: (mB+mA)a=mBg (m_B + m_A) \cdot a = m_B \cdot g
Solve for acceleration a a : a=mBgmA+mB a = \frac{m_B \cdot g}{m_A + m_B}
Substitute a a back into the equation for T T : T=mAmBgmA+mB T = m_A \cdot \frac{m_B \cdot g}{m_A + m_B}
The formulas for the acceleration of the system and the tension in the cord are: - Acceleration: a=mBgmA+mB a = \frac{m_B \cdot g}{m_A + m_B} - Tension: T=mAmBgmA+mB T = \frac{m_A \cdot m_B \cdot g}{m_A + m_B}

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