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Tension with a tension problem?

  • #1
"tension" with a tension problem?

11. [CJ6 4.P.076.] A 248 kg log is pulled up a ramp by means of a rope that is parallel to the surface of the ramp. The ramp is inclined at 29.0° with respect to the horizontal. The coefficient of kinetic friction between the log and the ramp is 0.880, and the log has an acceleration of 0.800 m/s2. Find the tension in the rope.


I dont know where to start please help me out!!!!!!!!!!!!!!!!!!!!!! :


surprised :cry: :surprised :cry:
 

Answers and Replies

  • #2
14
0
First of all, let us discuss the concept of the tension. Essentially, the tension is the force with which the block is being pulled. Therefore, the forces accross the "x" plane are the tension and the friction forces, as well as the "x" component of the gravitational force. On the "y" plane, we find the normal force as well as the "y" component of the gravitational force.
(note that I rotated our x-y plane by 29 degrees, so that all the forces are at 0 degrees to the surface )

Now, lets write what we've just said down:
[tex]F_{net,x} = F_t - F_k - F_{g,x}[/tex]
[tex]F_{net,y} = F_n - F_{g,y}[/tex]
To find [tex]F_k[/tex], we want to find the normal force. Since we know that no acceleration on the "y" plane exists (that is, the log doesn't levitate above the surface), [tex]F_{net,y} = ma = m(0) = 0[/tex]. So, [tex]F_n = F_{g,y}[/tex]. Looking at the block diagram you should've drawn, you should get [tex]F_{g,y}=F_n=mg sin{69^0}[/tex].
Let's go back to friction. [tex]F_k = \mu _k F_n = \mu _k (mg sin{69^0} )[/tex].
We substitute into our original [tex]F_{net,x}[/tex]:
[tex]m a_x = F_t - \mu _k (mg sin{69^0} ) - (mg cos{69^0})[/tex]
[tex] F_t = m a_x + \mu _k (mg sin{69^0} ) + (mg cos{69^0})[/tex]
Substitute, and you're done.

Hope this helps,
-Evgeny
 
  • #3
still tryin

Thanks for the help! I followed your steps above however the answer i got is not the correct answer. Am i doing the problem incorrect.

F=248kg(0.800m/s^2)+.880((248kg)(9.80m/s^2)(sin69))+248kg(9.80m/s^2)(cos69)
F=2367.23 is not correct
could you please tell me what i am doing wrong. :confused:
 
  • #4
298
0
Where did 69 come from? The ramp is on a 29 degree incline. Charvonne, i don't think you quite understand the problem, because although evgeny did make a mistake, his process is right. That is something you should be able to pick up on and figure out the solution.
 
  • #5
Doc Al
Mentor
44,869
1,116
Actually, Evgeny made two mistakes. One, he used 69 degrees instead of 29. But he also mixed up sine and cosine. For example, the normal force is [itex]mg cos\theta[/itex], not [itex]mg sin\theta[/itex]. Similarly, the component of the weight down the ramp is [itex]mg sin\theta[/itex], not [itex]mg cos\theta[/itex]. (Where [itex]\theta[/itex] = 29 degrees.)
 

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