Math  /  Algebra

QuestionYour answer is partially correct.
In the figure here three ballot boxes are connected by cords, one of which wraps over a pulley having negligible friction on its axle and negligible mass. The masses are mA=31 kg,mB=42 kgm_{A}=31 \mathrm{~kg}, m_{B}=42 \mathrm{~kg}, and mC=11 kgm_{C}=11 \mathrm{~kg}. When the assembly is released from rest, (a) what is the tension in the cord connecting B and C, and (b) how far does A move in the first 0.200 s (assuming it does not reach the pulley)? (a) Number \square Unit \square (b) Number \square ! Unit \square eTextbook and Media

Studdy Solution

STEP 1

What is this asking? We have three boxes connected by ropes and we want to find the tension in one of the ropes and how far one of the boxes moves in a short amount of time. Watch out! Don't forget to consider all the forces acting on each box!
Gravity pulls down, tension pulls up, and there's acceleration too.

STEP 2

1. Draw a free-body diagram for each box.
2. Find the system's acceleration.
3. Calculate the tension between boxes B and C.
4. Determine the distance box A moves.

STEP 3

Let's visualize the forces at play here!
Box A is hanging freely, experiencing a downward force due to gravity and an upward force due to the tension TABT_{AB} in the cord connecting it to box B.
Box B rests on a horizontal surface, with gravity pulling down, the normal force from the surface pushing up, TABT_{AB} pulling left, and the tension TBCT_{BC} from the cord connecting it to box C pulling to the right.
Finally, box C hangs freely, experiencing a downward force due to gravity and an upward force due to TBCT_{BC}.

STEP 4

Applying Newton's second law to each box, we get: For box A: mAgTAB=mAam_A \cdot g - T_{AB} = m_A \cdot a For box B: TBCTAB=mBaT_{BC} - T_{AB} = m_B \cdot a For box C: TBCmCg=mCaT_{BC} - m_C \cdot g = m_C \cdot a

STEP 5

We have three equations and three unknowns (TABT_{AB}, TBCT_{BC}, and aa).
Let's solve for the **system's acceleration**, aa.
Adding the three equations together, we get: mAgmCg=(mA+mB+mC)am_A \cdot g - m_C \cdot g = (m_A + m_B + m_C) \cdot a

STEP 6

Solving for aa, we have: a=(mAmC)gmA+mB+mCa = \frac{(m_A - m_C) \cdot g}{m_A + m_B + m_C} Substituting the given values (mA=31 kgm_A = \textbf{31 kg}, mB=42 kgm_B = \textbf{42 kg}, mC=11 kgm_C = \textbf{11 kg}, and g=9.8 m/s2g = \textbf{9.8 m/s}^2), we find: a=(31 kg11 kg)9.8 m/s231 kg+42 kg+11 kg=20 kg9.8 m/s284 kg2.33 m/s2a = \frac{(\textbf{31 kg} - \textbf{11 kg}) \cdot \textbf{9.8 m/s}^2}{\textbf{31 kg} + \textbf{42 kg} + \textbf{11 kg}} = \frac{\textbf{20 kg} \cdot \textbf{9.8 m/s}^2}{\textbf{84 kg}} \approx \textbf{2.33 m/s}^2

STEP 7

Now, let's find TBCT_{BC} using the equation for box C: TBC=mCa+mCg=mC(a+g)T_{BC} = m_C \cdot a + m_C \cdot g = m_C(a + g) Plugging in the values for mCm_C and the calculated aa, we get: TBC=11 kg(2.33 m/s2+9.8 m/s2)133 NT_{BC} = \textbf{11 kg}(\textbf{2.33 m/s}^2 + \textbf{9.8 m/s}^2) \approx \textbf{133 N}

STEP 8

Since box A starts from rest, we can use the equation for constant acceleration to find the distance it travels in the first 0.200 s\textbf{0.200 s}: d=v0t+12at2d = v_0 \cdot t + \frac{1}{2} \cdot a \cdot t^2 Since the **initial velocity** v0=0 m/sv_0 = \textbf{0 m/s}, the equation simplifies to: d=12at2d = \frac{1}{2} \cdot a \cdot t^2 Plugging in the calculated value of aa and t=0.200 st = \textbf{0.200 s}, we get: d=122.33 m/s2(0.200 s)20.0466 md = \frac{1}{2} \cdot \textbf{2.33 m/s}^2 \cdot (\textbf{0.200 s})^2 \approx \textbf{0.0466 m}

STEP 9

(a) The tension in the cord connecting B and C is approximately 133 N\textbf{133 N}. (b) Box A moves approximately 0.0466 m\textbf{0.0466 m} in the first 0.200 s\textbf{0.200 s}.

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