Net force of object indicated in following diagram

  • #1

Homework Statement


Calculate the net force acting on the object indicated in the following diagrams. Show your work.

1)
upload_2016-5-4_15-20-43.png


2)
upload_2016-5-4_15-21-18.png



Homework Equations


upload_2016-5-4_15-22-33.png

[/B]
upload_2016-5-4_15-22-57.png


The Attempt at a Solution



1)
8.0 N (north) + 10.0 N (south)
- 8.0 N (south) + 10.0 N (south)
= 2.0 N (south)

c= [ (2.0 N) 2 + (17.0 N) 2 - 2 (2.0 N) x (17.0 N) cos 45° ] 1/2
c= 15.6 N

sinA/2.0N = sin45°/15.6N

A= 5.2°
A= 45° - 5.2° = 39.8°

Fnet = 15.6 N [N 39.8° W]

2)
c= [ (12 N) 2 + (15 N)2 - 2 (12 N) x (15 N) cos124°]
c= 23.9 N

sinA/15N = sin124°/ 23.9 N

A = 31°

Fnet = 23.9 N [ N 31° E]

is this correct?
 

Answers and Replies

  • #2
haruspex
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I agree with both magnitudes.

In 1, you do not show a diagram for the triangle of forces, so I'm not sure what your angle A is. You seem to have used it inconsistently, leading to the wrong answer for direction of force.

In 2, the angle you found is not the one you want. It seems to be the angle between the resultant and the 12N force.

Can you post your force triangle diagrams?
 
  • #3
In 1, you do not show a diagram for the triangle of forces, so I'm not sure what your angle A is. You seem to have used it inconsistently, leading to the wrong answer for direction of force.
1)
Here is the diagram for 1) :

upload_2016-5-6_20-13-21.png

I corrected my answer for 1) :

8.0 N (north) + 10.0 N (south)

-8.0 N (south) + 10.0 N (south)

= 2.0 N (south)

c = [(2.0 N)2 + (17.0 N)2- 2 (2.0 N) x (17.0 N) cos 45°]1/2

c = 15.6 N

sinA/a = sinB/b = sinC/c

sinA/2.0 N = sin 45°/15.6 N

A = 5.2°

45° - 5.2° = 39.8°

Fnet = 16 N [ W 40° N]

is this better?
 
Last edited:
  • #4
In 2, the angle you found is not the one you want. It seems to be the angle between the resultant and the 12N force.
2)
Here is the diagram for 2):
upload_2016-5-6_20-26-14.png


I also corrected my answer for 2) :

c2 = a2 + b2 - 2abcosC

c = (a2 + b2 - 2abcosC)1/2

c = [(12 N)2 + (15 N)2- 2 (12 N) x (15 N) cos 124°]1/2

c = 23.9 N

sinA/a = sinB/b = sinC/c

sinA/15 N = sin 124°/ 23.9 N

A = 31°

32° - 31° = 1°

90° - 1° = 89°

Fnet = 24.0 N [N 89° E] or 24.0 N [E 1° N]

better?
 
Last edited:
  • #5
haruspex
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1)
Here is the diagram for 1) :

View attachment 100369
I corrected my answer for 1) :

8.0 N (north) + 10.0 N (south)

-8.0 N (south) + 10.0 N (south)

= 2.0 N (south)

c = [(2.0 N)2 + (17.0 N)2- 2 (2.0 N) x (17.0 N) cos 45°]1/2

c = 15.6 N

sinA/a = sinB/b = sinC/c

sinA/2.0 N = sin 45°/15.6 N

A = 5.2°

45° - 5.2° = 39.8°

Fnet = 16 N [ W 40° N]

is this better?
Looks right.

Edit: the angle might be a bit inaccurate. Retry it keeping more significant digits at each step.
Doing it a slightly different way (using arctan) I got about 42 degrees.
 
Last edited:
  • #6
haruspex
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2)
Here is the diagram for 2):
View attachment 100371

I also corrected my answer for 2) :

c2 = a2 + b2 - 2abcosC

c = (a2 + b2 - 2abcosC)1/2

c = [(12 N)2 + (15 N)2- 2 (12 N) x (15 N) cos 124°]1/2

c = 23.9 N

sinA/a = sinB/b = sinC/c

sinA/15 N = sin 124°/ 23.9 N

A = 31°

32° - 31° = 1°

90° - 1° = 89°

Fnet = 24.0 N [N 89° E] or 24.0 N [E 1° N]

better?
yes, but again the angle could be a touch more accurate.
 
  • #7
Edit: the angle might be a bit inaccurate. Retry it keeping more significant digits at each step.
Doing it a slightly different way (using arctan) I got about 42 degrees.
Ok, I will keep a few more significant digits to make the angle more accurate. Anyway, thanks for the help :)
 

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