Three blocks missing one mass

  • Thread starter bradsmith
  • Start date
  • #1
7
0
So I was sick last week for 3 classes and I really missed something and don't know what im doing.


Three blocks rest on a frictionless, horizontal table (see figure below), with m1 = 6 kg and m3 = 14 kg. A horizontal force F = 111 N is applied to block 1, and the acceleration of all three blocks is found to be 3.4 m/s2

(a) Find m2.

For this I got M2=12kg It was wrong
I did F=ma for both blocks
F=6x3.4=20.4
F=14x3.4=47.6
111-20.4-47.6=42.6
42.6/3.4=12

I think I have to use gravity 9.8?

(b) What is the magnitude of the normal force between blocks 2 and 3?
 

Answers and Replies

  • #2
tjmiller88
Gold Member
13
2
Hi bradsmith,

For part (a), because the 3 blocks are touching each other, they really just act like one mass, so the way you are splitting up the forces is not correct. Your formula should actually be F=(m1+m2+m3)*a. Make sense?

You would only have to use gravity in this type of problem if there was friction involved, but in this case there is none.

Now for part (b), you want to ask yourself what force needs to be applied to block 3 in order for it to travel 3.4 m/s2.

See if that helps.
 
  • #3
35,513
11,972
Hi bradsmith,

For part (a), because the 3 blocks are touching each other, they really just act like one mass, so the way you are splitting up the forces is not correct. Your formula should actually be F=(m1+m2+m3)*a. Make sense?
It is no problem to split up the forces like that.

The problem is just in the calculations. bradsmith: check your numbers and you'll find the error.
 
  • #4
7
0
Ok F=(m1+m2+m3)xa
111=(6+m2+14)X3.4
111=(m2+20)x3.4
111=3.4m+68
111-68=3.4m
43/3.4m=12.64

is b 90.57
F=a(m2+m3)
F=3.4(12.64+14)
F=90.57
Thanks for not giving me the answer, I learn better by doing it.
I think I did something wrong again
 
  • #5
35,513
11,972
Why did you add masses 2 and 3 for (b)?
You didn't attach the figure, but I guess the masses are in the order "1 2 3" and the force acts at block 1. So the force between 2 and 3 just have to accelerate block 3.

Thanks for not giving me the answer, I learn better by doing it.
That's a forum rule for exactly this reason :).
 
  • #6
7
0
Yes it force is applied to 1 and is in order as 123. so that means add all 3
6+12.64+14+32.64
f/m=a
111/32.64=3.4
so its 32.64?

Im not sure why I did it that way. I thought because I only needed 2 and 3 that was the way to do it.
 
  • #7
35,513
11,972
What is 32.64 now?
And what are its units?
I thought because I only needed 2 and 3 that was the way to do it.
What do you mean with "needed"?
 
  • #8
7
0
Let me start over.
Three blocks rest on a frictionless, horizontal table (see figure below), with m1 = 6 kg and m3 = 14 kg. A horizontal force F = 111 N is applied to block 1, and the acceleration of all three blocks is found to be 3.4 m/s2

all 3 blocks are side by side F--->123

(a) Find m2

F=(m1+m2+m3)xa
111N=(6kg+m2+14kg)X3.4m/s2
111N=(m2+20kg)x3.4m/s2
111N=3.4m+68kg
111N-68=3.4m/s2
43N/3.4m/s2=12.64kg

(b) What is the magnitude of the normal force between blocks 2 and 3?



F=ma
111=(m1+m2)a
111=26.64*3.4
111/90.57=1.22


I don't know what the other stuff is from
 
Last edited:
  • #9
35,513
11,972
F=ma
111=(m1+m2)a
Is that true? Do you use a force of 111 N just to accelerate blocks 1 and 2?
111=26.64*3.4
3=5?
111/90.57=1.22
Randomly mixing numbers does not help.

Forget about blocks 1 and 2 for a while. Which forces act on block 3? What acceleration results from those forces?
 
  • #10
7
0
NVM I got it. F=ma
F=14kg*3.4m/s2
F=47.6

Thanks everyone.
 

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