- #1
Dorothy Weglend
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- 2
This is problem 5-61 in Serway and Jewett, 4th Ed. (See attached figure).
The problem is to find the force F which will keep the blocks stationary relative to the cart. All surfaces, wheels and pulley are frictionless.
I get F=(m1+m2+M)(g*m2/(m1+m2)
The answer in the back of the book is
F=(m1+m2+M)(g*m2/m1)
Is this right? Here is my work (T is tension in the string):
T - m1*a = 0
m2*g - T = m2*a
Solving for a: a = g*m2/(m1+m2)
The solution in the book must have come from these two equations:
T - m1*a = 0
m2*g - T = 0
which are from the point of view of an observer outside of the cart, I guess.
I realize that if my solution works, the books will also, since it gives a larger F, but will the smaller acceleration not be enough?
Perhaps I don't understand inertial frames very well yet.
Thanks,
Dorothy
The problem is to find the force F which will keep the blocks stationary relative to the cart. All surfaces, wheels and pulley are frictionless.
I get F=(m1+m2+M)(g*m2/(m1+m2)
The answer in the back of the book is
F=(m1+m2+M)(g*m2/m1)
Is this right? Here is my work (T is tension in the string):
T - m1*a = 0
m2*g - T = m2*a
Solving for a: a = g*m2/(m1+m2)
The solution in the book must have come from these two equations:
T - m1*a = 0
m2*g - T = 0
which are from the point of view of an observer outside of the cart, I guess.
I realize that if my solution works, the books will also, since it gives a larger F, but will the smaller acceleration not be enough?
Perhaps I don't understand inertial frames very well yet.
Thanks,
Dorothy