Math Problem Statement

What is the magnitude and direction of the acceleration of each block?

What is the magnitude and direction of the acceleration of each block?

Solution

To solve for the magnitude and direction of the acceleration of each block in this pulley system, we need to:

  1. Apply Newton's second law to each mass, considering forces acting on them.
  2. Consider the system's constraints (acceleration and forces across pulleys and tension in the strings).
  3. Set up a system of equations to solve for acceleration aa.

Step-by-Step Analysis:

We are given:

  • mA=1.25kgm_A = 1.25 \, \text{kg}
  • mB=2.50kgm_B = 2.50 \, \text{kg}
  • mC=3.75kgm_C = 3.75 \, \text{kg}
  • Coefficient of kinetic friction μk=0.602\mu_k = 0.602 for block mBm_B.
  • All blocks accelerate with the same magnitude of acceleration aa.

1. Forces on Each Block:

  • Block A: Since it's hanging, the forces acting are:

    • Gravitational force: FgA=mAgF_{gA} = m_A g
    • Tension in the string: TAT_A

    Applying Newton's second law: m_A g - T_A = m_A a \tag{1}

  • Block B: This block is on the ground, with friction acting on it. Forces are:

    • Gravitational force: FgB=mBgF_{gB} = m_B g (normal force)
    • Frictional force: Ffriction=μkFgB=μkmBgF_{\text{friction}} = \mu_k F_{gB} = \mu_k m_B g
    • Tension in the string pulling: TBT_B

    Newton's second law for block BB: T_B - \mu_k m_B g = m_B a \tag{2}

  • Block C: Another hanging mass, with forces:

    • Gravitational force: FgC=mCgF_{gC} = m_C g
    • Tension in the string: TCT_C

    Newton's second law for block CC: T_C - m_C g = -m_C a \tag{3} (Note that CC accelerates downward, hence the negative sign).

2. System Constraints:

For an ideal pulley system, we can assume that the tension in the string is the same across the pulleys (due to massless and frictionless assumptions). Therefore, we have TA=TB=TCT_A = T_B = T_C.

3. Solving the System:

We now have three equations:

  1. mAgT=mAam_A g - T = m_A a
  2. TμkmBg=mBaT - \mu_k m_B g = m_B a
  3. TmCg=mCaT - m_C g = -m_C a

Solving these equations step by step:

  • From equation (3): T=mCgmCaT = m_C g - m_C a

  • Substituting TT into equation (1): mAg(mCgmCa)=mAam_A g - (m_C g - m_C a) = m_A a Simplifying: mAgmCg+mCa=mAam_A g - m_C g + m_C a = m_A a mAgmCg=(mA+mC)am_A g - m_C g = (m_A + m_C) a a=mAgmCgmA+mCa = \frac{m_A g - m_C g}{m_A + m_C} Substituting values mA=1.25kgm_A = 1.25 \, \text{kg}, mC=3.75kgm_C = 3.75 \, \text{kg}, and g=9.8m/s2g = 9.8 \, \text{m/s}^2: a=(1.25)(9.8)(3.75)(9.8)1.25+3.75a = \frac{(1.25)(9.8) - (3.75)(9.8)}{1.25 + 3.75} a=12.2536.755=24.55=4.9m/s2a = \frac{12.25 - 36.75}{5} = \frac{-24.5}{5} = -4.9 \, \text{m/s}^2

This value implies that block CC moves downward, and block AA moves upward with an acceleration of 4.9m/s24.9 \, \text{m/s}^2.

Would you like further details or explanations on any part of this solution? Here are some related questions that might help clarify:

  1. How does tension in the string affect the system?
  2. Why is the acceleration the same for all blocks?
  3. How would the result change if the coefficient of friction were different?
  4. What happens if the pulleys were not frictionless?
  5. How does Newton's second law apply to each block individually?

Tip: When analyzing pulley systems, always write out the force equations for each object involved and consider the direction of forces and accelerations carefully.

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Math Problem Analysis

Mathematical Concepts

Newton's Second Law of Motion
Friction
Pulley System Dynamics

Formulas

F = ma
F_friction = μ_k * N
T - mC * g = -mC * a
mA * g - T = mA * a
T - μ_k * mB * g = mB * a

Theorems

Newton's Second Law

Suitable Grade Level

High School Physics (Grades 10-12)