Math Problem Statement

The system shown in the figure is initially kept at rest. The mass of the object A in the picture is 24 kg and the mass of the object B is 25 kg. Let the coefficient of static friction between A and B be 0.26 and the coefficient of kinetic friction between B and the surface be 0.13. When C is released, what is the maximum mass m_C that C can have to cause A and B to slide together? You can assume that the string is massless and slides smoothly over the pulley. a) Calculate the maximum mass of the piece C. b) Calculate the frictional force between object B and the surface.

Solution

Let's solve the problem step by step.

Part (a) Maximum mass of piece C (mCm_C)

We are asked to calculate the maximum mass of CC that can cause both blocks AA and BB to move together without slipping. To solve this, we'll apply Newton's laws and frictional forces.

  1. Forces acting on block A (on top of B):

    • Weight of AA: WA=mAg=24kg9.8m/s2=235.2NW_A = m_A \cdot g = 24 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 235.2 \, \text{N}

    • The friction between AA and BB must prevent AA from slipping when BB moves. The static friction force is: fs=μsNAf_s = \mu_s \cdot N_A where μs=0.26\mu_s = 0.26 is the coefficient of static friction and NA=WA=235.2NN_A = W_A = 235.2 \, \text{N}.

      Thus, fs=0.26235.2=61.15Nf_s = 0.26 \cdot 235.2 = 61.15 \, \text{N}

  2. Forces acting on block B (on the surface):

    • The total weight on BB is the combined mass of AA and BB: WB=(mA+mB)g=(24+25)9.8=499.8=480.2NW_B = (m_A + m_B) \cdot g = (24 + 25) \cdot 9.8 = 49 \cdot 9.8 = 480.2 \, \text{N}
    • The kinetic friction between BB and the surface is: fk=μkNB=0.13480.2=62.43Nf_k = \mu_k \cdot N_B = 0.13 \cdot 480.2 = 62.43 \, \text{N}
  3. For block CC:

    • The force due to the weight of CC is: FC=mCgF_C = m_C \cdot g
    • For the blocks to move together, the force on BB (caused by CC) must overcome both the kinetic friction between BB and the surface and the static friction between AA and BB.

    Therefore, the force balance for BB gives: mCg=fs+fk=61.15+62.43=123.58Nm_C \cdot g = f_s + f_k = 61.15 + 62.43 = 123.58 \, \text{N} mC=123.589.8=12.61kgm_C = \frac{123.58}{9.8} = 12.61 \, \text{kg}

Thus, the maximum mass of CC is 12.61 kg.


Part (b) Frictional force between object B and the surface

As calculated earlier, the frictional force between BB and the surface is the kinetic friction force: fk=62.43Nf_k = 62.43 \, \text{N}

Summary:

  • (a) The maximum mass of piece CC is 12.61kg12.61 \, \text{kg}.
  • (b) The frictional force between object BB and the surface is 62.43N62.43 \, \text{N}.

Would you like more details on any part of this explanation or to clarify the concepts used?


Here are some related questions you might find helpful:

  1. How does the coefficient of friction affect the force needed to move objects?
  2. What would happen if the coefficient of kinetic friction was larger than the static friction?
  3. What role does normal force play in determining frictional forces?
  4. How does the pulley system affect the motion of the blocks?
  5. Could the mass of CC be greater if the surface had less friction?

Tip: Static friction keeps objects from moving, and it is generally stronger than kinetic friction, which acts when objects are already in motion.

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Math Problem Analysis

Mathematical Concepts

Classical Mechanics
Friction
Newton's Laws
Force Balance

Formulas

f_s = μ_s * N_A (static friction)
f_k = μ_k * N_B (kinetic friction)
m_C * g = f_s + f_k (force balance)

Theorems

Newton's Second Law of Motion

Suitable Grade Level

Grades 10-12