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Math  /  Algebra

Question58. (III) Suppose the pulley in Fig. 4-50 is suspended by a cord C. Determine the tension in this cord after the masses are released and before one hits the ground. Ignore the mass of the pulley and cords.

Studdy Solution

STEP 1

1. The pulley is massless and frictionless.
2. The cords are massless and inextensible.
3. The system is in a uniform gravitational field with acceleration due to gravity g=9.8 m/s2 g = 9.8 \, \text{m/s}^2 .
4. The tension in the cord C is the same throughout.
5. We are trying to find the tension in cord C after the masses are released and before one hits the ground.

STEP 2

1. Analyze the forces acting on each mass.
2. Write the equations of motion for each mass.
3. Solve for the acceleration of the system.
4. Determine the tension in the cords connecting the masses.
5. Calculate the tension in cord C.

STEP 3

Analyze the forces acting on each mass.
For the left mass (m1=2.2 kg m_1 = 2.2 \, \text{kg} ): - Weight W1=m1⋅g=2.2⋅9.8 N W_1 = m_1 \cdot g = 2.2 \cdot 9.8 \, \text{N} - Tension T1 T_1 in the cord
For the right mass (m2=4.2 kg m_2 = 4.2 \, \text{kg} ): - Weight W2=m2⋅g=4.2⋅9.8 N W_2 = m_2 \cdot g = 4.2 \cdot 9.8 \, \text{N} - Tension T2 T_2 in the cord

STEP 4

Write the equations of motion for each mass.
For the left mass (m1 m_1 ): m1⋅a=T1−W1 m_1 \cdot a = T_1 - W_1
For the right mass (m2 m_2 ): m2⋅a=W2−T2 m_2 \cdot a = W_2 - T_2
Since the pulley is massless and frictionless, T1=T2=T T_1 = T_2 = T .

STEP 5

Solve for the acceleration of the system.
Add the two equations: m1⋅a+m2⋅a=W2−W1 m_1 \cdot a + m_2 \cdot a = W_2 - W_1
(m1+m2)⋅a=m2⋅g−m1⋅g (m_1 + m_2) \cdot a = m_2 \cdot g - m_1 \cdot g
a=(m2−m1)⋅gm1+m2 a = \frac{(m_2 - m_1) \cdot g}{m_1 + m_2}
Substitute the values: a=(4.2−2.2)⋅9.82.2+4.2 a = \frac{(4.2 - 2.2) \cdot 9.8}{2.2 + 4.2}
a=2⋅9.86.4 a = \frac{2 \cdot 9.8}{6.4}
a=19.66.4 a = \frac{19.6}{6.4}
a=3.0625 m/s2 a = 3.0625 \, \text{m/s}^2

STEP 6

Determine the tension in the cords connecting the masses.
Using the equation for the right mass: m2⋅a=W2−T m_2 \cdot a = W_2 - T
T=W2−m2⋅a T = W_2 - m_2 \cdot a
T=4.2⋅9.8−4.2⋅3.0625 T = 4.2 \cdot 9.8 - 4.2 \cdot 3.0625
T=41.16−12.8625 T = 41.16 - 12.8625
T=28.2975 N T = 28.2975 \, \text{N}

STEP 7

Calculate the tension in cord C.
Since the pulley is massless and the tension in the cords on either side is equal, the tension in cord C is the sum of the tensions in the cords connecting the masses:
TC=2T T_C = 2T
TC=2×28.2975 T_C = 2 \times 28.2975
TC=56.595 N T_C = 56.595 \, \text{N}
The tension in cord C is:
56.595 N \boxed{56.595 \, \text{N}}

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