There were two questions concerning the forces acting on the carts connected by

Question

There were two questions concerning the forces acting on the carts connected by a string that you were to ponder: Which object, the string (between the carts) or cart A, exerts a force directly on cart B? Which object, the string or cart A, has a force acting directly on it from cart B? In light of your answers to the above question, do the forces measured in steps 2.1 through 2.4 (that is, the forces registered by each force sensor during a Pull-Hold-Release sequence) qualify as a "third law pair"? If these forces are not third-law pairs, why would they be equal in magnitude (within the uncertainty in the measurement)? The following hypothetical experiment will help you think about these questions. Imagine that instead of a string between the two carts, you have instead a "magic cart" whose mass m2 can be varied at will, all the way to zero: Assume the tension in the string is T = 2.00 N, m_1 = 2.00 kg and m_3 = 3.00 kg. Also assume that friction can be neglected. Get a piece of paper and pencil and use the three free body diagrams drawn above the carts to find equations for all of the forces in terms of m_1, m_2, m_3 and T. You can use Newton's second law to relate the forces and the acceleration of each cart, which is the same for all of the carts, and you can use the third law to relate the magnitudes of the third law pairs. From these equations calculate the values of the four forces for four different values of the mass of the magic cart (including 0 kg), and fill in the table below. Signs are important! Assume that the positive direction is to the right. Magnitudes must be correct to within 1% of the key. You should put in three significant figures to ensure that you are within tolerance. After you fill in the table with the correct values (no red hints!), what do you notice about F_12 and F_32 when m_2 gets smaller? In what way is a string like the magic cart?

Explanation / Answer

Step 2.6
a) String between the carts exwrts the force directly on the cart
b) String between the carts exerts the force directly on the cart
c) These force would qualify as the third law pair ( even if they are connected indirectly)

T = 2 N
m1 = 2 kg
m3 = 3 kg

T - F21 = m1 *a
F12 - F32 = m2 * a
F23 = m3*a

adding 1 and 3

T - (m2 * a) = (m1 + m2)*a
T = (m1 + m2 + m3)*a
F21 = F12 = (m1 + m2 + m3)*a - m1*a = (m2 + m3)a
F23 = F32 = m3*a

now, T = 2N, m1 = 2 kg, m3 = 3 kg

assume, m2 = 0 kg
T = 5a = 2, a = 0.4 m/s/s
T = 2 N
F21 = F12 = 1.2 N
F23 = F32 = 1.2 N

Assume m2 = 1 kg
T = 6a = 2, a = 1/3 m/s/s
T = 2 N
F21 = F12 = 4/3 N
F23 = F32 = 1 N

aSSUME m2 = 2 kg
T = 7a = 2, a = 2/7 m/s/s
T = 2N
F21 = F12 = 10/7 N
F23 = F32 = 6/7 N

F21 = F32 as soon as the mass of magic cart dissaperas

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