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Pullign Blocks Question

  • Thread starter Paymemoney
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Homework Statement


Three identical blocks connected by ideal strings are being pulled along a horizontal frictionless surface by a horizontal force F_vec. The magnitude of the tension in the string between blocks B and C is [tex]T = 3.00 {\rm N}[/tex]. Assume that each block has mass [tex]m = 0.400 {\rm kg}[/tex].

a)What is the magnitude F of the force?

b) What is the tension T_AB in the string between block A and block B?

http://img340.imageshack.us/img340/3789/9374t.jpg [Broken]


Homework Equations


W=mg
F=ma


The Attempt at a Solution


a) i made A and B as one object and calculated the acceleration of the object

two forces are acting on this object.
1st force
w=mg
w=0.4*9.8
W=3.92N

2nd force is tension = 3N

Found Fnet to be 3+3.92+4.94
Fnet = 11.86

used F=ma formula

a=11.86/0.4
a=29.65m/s^2

then to find the force i went: a=Fnet/3m

so a=11.86/3*0.4

a=9.88m/s

To find force i use 9.88 * 0.4 = 4N ------> answer's says it 4.50N (tell me where i have gone wrong)

b) Used f=ma

f= 29.65 * 0.4
=11.86 i know this is incorrect, but someone tell me the correct way of doing this.

P.S
 
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Answers and Replies

  • #2
Doc Al
Mentor
44,877
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a) i made A and B as one object and calculated the acceleration of the object

two forces are acting on this object.
1st force
w=mg
w=0.4*9.8
W=3.92N

2nd force is tension = 3N
There's also the upward normal force acting on the blocks, which exactly cancels the weight. (Since there's no acceleration in the vertical direction.)

Found Fnet to be 3+3.92+4.94
Fnet = 11.86
I thought you said there were two forces acting on the blocks? And since these forces act in different directions, they can't just be added like numbers.

Hint: The vertical forces add to zero, so they can be ignored.
 

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