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A Sliding Block

  • Thread starter cd.riter
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  • #1
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Block 1 as shown in the link below(url) has a mass of 10.3 kg. The coefficient of static friction between the block and the table is μs = 0.36 and the coefficient of kinetic friction is μk = 0.336.

a) What should the mass of the second block be so that m1 is just on the verge of sliding?
I ended up getting (10.3 kg)*(0.36) , which was correct

b) If the block 2 has the mass you just calculated above, what is the tension in the string?
I got (10.3*0.36)(9.81), and that was also correct

c) If the block 2 has the mass you just calculated above and then block 1 is given a slight nudge, what is the tension in the string?
So i did the FBD, to solve for acceleration of the blocks, and i got ((M2 - Uk*M1)/(M1+M2)) *g = a
i plugged acceleration back into this equation Ft = m1*a + Ffric => where Ffric is (Uk*m1*g)
and that was not correct please help me

here's the pic
https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/phys2111/fall/homework/Ch-06-Forces/tipler_sliding_block/fig_05_10.jpg [Broken]
 
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Answers and Replies

  • #2
Doc Al
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c) If the block 2 has the mass you just calculated above and then block 1 is given a slight nudge, what is the tension in the string?
So i did the FBD, to solve for acceleration of the blocks, and i got ((M2 - Uk*M1)/(M1+M2)) *g = a
i plugged acceleration back into this equation Ft = m1*a + Ffric => where Ffric is (Uk*m1*g)
and that was not correct please help me
Looks fine to me. Double check your arithmetic.
 
  • #3
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yea i looked over everything many times, I'm beginning to think that its the computer program that's wrong...Well if anything my answer for Ft = 35.73 N, if you are getting anything else please let me know
 
  • #4
Doc Al
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yea i looked over everything many times, I'm beginning to think that its the computer program that's wrong...Well if anything my answer for Ft = 35.73 N, if you are getting anything else please let me know
What did you get for the acceleration? (Often the online programs are fussy about significant figures.)
 
  • #5
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i got .173 m/s/s
so i put in the calculator

(10.3)(.173) + (.336*10.3*9.81) = Ft
 
  • #6
Doc Al
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i got .173 m/s/s
Seems a bit off--by a factor of 10. Double check this.
 
  • #7
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I can't find my error, thanks anyway Doc, I'll just go find my professor, I'm sure he's made an error in programming this, he's done it before
 
  • #8
Doc Al
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I can't find my error, thanks anyway Doc, I'll just go find my professor, I'm sure he's made an error in programming this, he's done it before
Before you go to your professor, show the details of your calculation of the acceleration.
 
  • #9
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first I made the tensions equal to each other

m1*a + Uk*m1*g = m2g - m2 a , then i solved for a

a = ((m2 - uk*m1)/(m1+m2)) * g


a = (((10.3*.36)-(.336*10.3)/((10.3*.36)+10.3))) * 9.81 = .173
 
  • #10
Doc Al
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first I made the tensions equal to each other

m1*a + Uk*m1*g = m2g - m2 a , then i solved for a

a = ((m2 - uk*m1)/(m1+m2)) * g


a = (((10.3*.36)-(.336*10.3)/((10.3*.36)+10.3))) * 9.81 = .173
Yep, looks OK. (I misread something earlier.)
 

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