Forces with angles, Find the acceleration

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 3K views
plumeria28
Messages
4
Reaction score
0

Homework Statement



Two forces act on a 4.00 kg mass which is sitting at rest on a horizontal frictionless surface. One force is 6.50 N directed 55° West of South, the other is 7.00 N directed 25° North of West. What acceleration does the mass receive?

Homework Equations



sin = o/h

cos = a/h

tan = o/a

a = F/m

The Attempt at a Solution



I drew forces (not accurately) on paint (attachment) to help me solve this question then did the following calculations:

Fapx1
sin = o/h sin35° = o/6.5N o = 3.7N

Fapx2
sin = o/h sin25° = o/7N o = 3N [N]

Fapx = Fapx1 + Fapx2
Fapx = 3.7N - 3N
Fapx = 0.7N

Fapy1
cos = a/h cos35° = a/6.5N a = 5.3N [W]

Fapy2
cos = a/h cos25° = a/7N a = 6.3N [W]

Fapy = Fapy1 + Fapy2
Fapy = 5.3N + 6.3N
Fapy = 11.6N

************************************************
tan = o/a tan = 11.6/0.7 tan = 86.5°

sin = o/h sin86.5° = 11.6/h h = 11.622N

************************************************

a = F/m a = 11.622N/4.00kg a = 2.9 m/s^2
_____________________________________________________________________________
I was wondering if I did the question right. Did I get the right answer?
 

Attachments

  • 7.jpg
    7.jpg
    16.3 KB · Views: 514
Physics news on Phys.org
plumeria28 said:

Homework Statement



Two forces act on a 4.00 kg mass which is sitting at rest on a horizontal frictionless surface. One force is 6.50 N directed 55° West of South, the other is 7.00 N directed 25° North of West. What acceleration does the mass receive?

Homework Equations



sin = o/h

cos = a/h

tan = o/a

a = F/m

The Attempt at a Solution



I drew forces (not accurately) on paint (attachment) to help me solve this question then did the following calculations:

Fapx1
sin = o/h sin35° = o/6.5N o = 3.7N

Fapx2
sin = o/h sin25° = o/7N o = 3N [N]

Fapx = Fapx1 + Fapx2
Fapx = 3.7N - 3N
Fapx = 0.7N

Fapy1
cos = a/h cos35° = a/6.5N a = 5.3N [W]

Fapy2
cos = a/h cos25° = a/7N a = 6.3N [W]

Fapy = Fapy1 + Fapy2
Fapy = 5.3N + 6.3N
Fapy = 11.6N

************************************************
tan = o/a tan = 11.6/0.7 tan = 86.5°

sin = o/h sin86.5° = 11.6/h h = 11.622N

************************************************

a = F/m a = 11.622N/4.00kg a = 2.9 m/s^2
_____________________________________________________________________________
I was wondering if I did the question right. Did I get the right answer?

Homework Statement


Homework Equations


The Attempt at a Solution



problems:

Firstly, you have not helped yourself with your diagram.

"Your" 55o angle is smaller than your 25o angle, so when you got two x-components of 3N and 3.7N you did not recognise that at least one of them was wrong, as the x-component of the 6.5N force is way less than the x component of the 7N force. The Y components don't llok good either. You seem to have added the y components and subtracted the x ??
 
Last edited:
I said the diagrams are not accurate, I just drew them so it would be visually easier to do the calculations. And I added the y components because they were both going in the same direction (west) and subtracted the x components because they were going opposite directions (south and north).
 
plumeria28 said:
I said the diagrams are not accurate, I just drew them so it would be visually easier to do the calculations. And I added the y components because they were both going in the same direction (west) and subtracted the x components because they were going opposite directions (south and north).

But your diagram clearly shows the x components both going west, and the y-components going in opposite directions - North and South.

You need to draw a more accurate diagram and stick to the conventions you state on that diagram if you want to get anywhere with this problem.