Math  /  Algebra

Question23. (II) In Fig. 5-36 the coefficient of static friction between mass mAm_{\mathrm{A}} and the table is 0.40 , whereas the coefficient of kinetic friction is 0.30 . (a) What minimum value of mAm_{\mathrm{A}} will keep the system from starting to move? (b) What value(s) of mAm_{\mathrm{A}} will keep the system moving at constant speed? [Ignore masses of the cord and the (frictionless) pulley.]

Studdy Solution

STEP 1

1. The system consists of two masses, mA m_A on a table and mB=2.0kg m_B = 2.0 \, \text{kg} hanging vertically.
2. The coefficient of static friction between mA m_A and the table is 0.40.
3. The coefficient of kinetic friction between mA m_A and the table is 0.30.
4. The pulley is frictionless, and the cord's mass is negligible.
5. We need to find the minimum value of mA m_A to prevent motion and the value(s) of mA m_A for constant speed.

STEP 2

1. Analyze the forces acting on the system.
2. Calculate the minimum mA m_A to prevent motion using static friction.
3. Calculate mA m_A for constant speed using kinetic friction.

STEP 3

Analyze the forces acting on the system.
- For mA m_A : The forces are tension T T , static or kinetic friction f f , and gravitational force mAg m_A g . - For mB m_B : The forces are tension T T and gravitational force mBg m_B g .

STEP 4

Calculate the minimum mA m_A to prevent motion using static friction.
- The static friction force fs f_s is given by fs=μsN f_s = \mu_s \cdot N , where N=mAg N = m_A g is the normal force. - The system will not move if the static friction force can balance the weight of mB m_B , i.e., fsmBg f_s \geq m_B g .
μsmAgmBg \mu_s \cdot m_A g \geq m_B g
- Simplify to find mA m_A :
mAmBμs m_A \geq \frac{m_B}{\mu_s}
Substitute the known values:
mA2.0kg0.40 m_A \geq \frac{2.0 \, \text{kg}}{0.40}
mA5.0kg m_A \geq 5.0 \, \text{kg}

STEP 5

Calculate mA m_A for constant speed using kinetic friction.
- For constant speed, the kinetic friction force fk f_k balances the weight of mB m_B , i.e., fk=mBg f_k = m_B g . - The kinetic friction force fk f_k is given by fk=μkN f_k = \mu_k \cdot N , where N=mAg N = m_A g .
μkmAg=mBg \mu_k \cdot m_A g = m_B g
- Simplify to find mA m_A :
mA=mBμk m_A = \frac{m_B}{\mu_k}
Substitute the known values:
mA=2.0kg0.30 m_A = \frac{2.0 \, \text{kg}}{0.30}
mA6.67kg m_A \approx 6.67 \, \text{kg}
The minimum value of mA m_A to prevent motion is 5.0kg 5.0 \, \text{kg} , and the value of mA m_A to keep the system moving at constant speed is approximately 6.67kg 6.67 \, \text{kg} .

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