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Acceleration of the block

  • #1
1. What is the acceleration of the 20 kg block if the block is being pulled with a force of 150 N at 60 degrees above the X+ axis and there is a frictional force of 20 N acting on the block?



2. F=ma, Fcos(theta).



3. I did (150)cos(60)=(20)a ---> 75=20a ---> a=3.75 m/s

Just wondering if this was correct and if I did it the right way. Thank you
 

Answers and Replies

  • #2
You have to remember that the friction force also applies here.

[itex]\Sigma[/itex]F = m*a

[itex]\Sigma[/itex]F = Fcos(theta) - friction = ma

Correct?
 
  • #3
16
0
I don't think you used the frictional force of 20N in your equation.

You have to remember that the friction force also applies here.

[itex]\Sigma[/itex]F = m*a

[itex]\Sigma[/itex]F = Fcos(theta) - friction = ma

Correct?
Wouldn't it be [itex]\Sigma[/itex]F = Fnetcos(theta) = ma
 
  • #4
I don't think you used the frictional force of 20N in your equation.



Wouldn't it be [itex]\Sigma[/itex]F = Fnetcos(theta) = ma
You wouldn't multiply the frictional force by cos(theta) though.

If you draw your FBD....

You have a friction force going straight backwards with a magnitude of 20 N. Therefore, it has an angle of 180 with direction of motion, so cos(180) = -1. The 60.0 degrees ONLY affects the diagonal force.

[itex]\Sigma[/itex]F = Fcos(theta) - friction = ma

:smile:
 
  • #5
16
0
You wouldn't multiply the frictional force by cos(theta) though.

If you draw your FBD....

You have a friction force going straight backwards with a magnitude of 20 N. Therefore, it has an angle of 180 with direction of motion, so cos(180) = -1. The 60.0 degrees ONLY affects the diagonal force.

[itex]\Sigma[/itex]F = Fcos(theta) - friction = ma

:smile:
How do you know that the angle of the friction is 180? Wouldn't the angle of friction be 240 degrees, because of the 60 degree incline?

How did you draw your FBD?
 
  • #6
How do you know that the angle of the friction is 180? Wouldn't the angle of friction be 240 degrees, because of the 60 degree incline?
Consider that the Force being applied is someone pulling on a string. That string is 60.0 degrees to the flat surface the block is on. Ergo, the Force is at 60.0 degrees to the horizontal. And the friction is acting on the flat surface, keeping the angle at 180.


But, to answer your question about a 60 degree incline.....

Even if the block was on a 60 degree incline, the friction force's angle would still be 180 degrees. Since the direction of motion would still be going up the ramp, the direction of friction would be going the exact opposite of the direction of motion, making a 180 degree angle (straight line).

:smile:

I drew my FBD:

Weight force going down.
Normal force going up.
Friction force going to the right.
Some force going up and to the left, forming a 60.0 angle with the horizontal.

I assume the block is on a flat surface.
 
Last edited:
  • #7
16
0
Oh, I assumed the block was on a ramp. That's why I was getting confused.
 

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