Could someone explain these steps to me, Im pretty sure i have me free body diag
ID: 1327233 • Letter: C
Question
Could someone explain these steps to me, Im pretty sure i have me free body diagram is correct, but i think im confused on the effects of the forces. Just need some clarity.
draw free body diagrams and use them to explain your reasoning in each part The figure at right shows two carts, connected by ) The following is a qualitative thought experiment. To receive full credit, you must a lightweight string. Another string runs over a massless, frictionless pulley to a hanging mass Cart A is held is held stationary by a hand. Compare the tension in string I to the tension in string 2? Be explicit about which of Newton's laws applies. Explain with a free body diagram. Find an expression for the tension. (Hint: for which object should you draw a free body diagram?) Now cart A is released and the system accelerates to the right. Note: The masses of A and B are not known. Compare the acceleration of Cart A to Cart B. Explain. (Hint: the string cannot stretch. No free body diagram needed here.) Compare the tension in string I to the tension in string 2 as the system accelerates. Explain. (Hint: for which object should you draw a free body diagram?) Eyplain Hint for wb shod you draw a free body diagram?) Compare the tension in string 2 as the system accelerates to its value when the system is held stationary. Explain. (Hint: for which object should you draw a free body diagram?) Compare the tension in string 1 as the system accelerates to its value when the system is held stationary. Explain. (Hint: you have already drawn the free body diagrams you need to answer this.)Explanation / Answer
When held stationary , a = 0
tension in 1 = T' and in 2 = T
balancing force along surface on Cart B
Fnet = ma
T - T' = ma = 0
so T = T'
that means Tension in both string is same.
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Balancing force on mass m in vertical ,
Fnet = ma
mg - T = ma = 0
T = mg
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now when cart is released, there will be some acceleration a.
Cart A and Cart B are connected by a rope that can't stretch so acc. of both cart will same.
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on block B,
T - T' = (mB)a
so T - T' > 0
T > T'
tenaion in string 2 > Tension in string 1
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When stationary T = T' = mg
when acc. is there
mg - T = ma
T = mg - ma < mg
in case of acc. Tension is less than the case of held stationary.
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When stationary T' = mg
When acc.
On block B,
T - T' = ma
T - ma = T'
T' = mg - ma
in case of acc. Tension is less than the case of held stationary.