# Net force (Diagrams incl) Please check my solutions

1. Mar 2, 2012

### supernova1203

Net force (Diagrams incl) Please check my solutions :)

1. The problem statement, all variables and given/known data
Calculate the net force acting on each object indicated in the following diagrams. Show your work.

The diagrams are in attachment, you dont have to download them to view them :)

2. Relevant equations
c^2=a^2+b^2-2abCosC

c^2=√a^2+b^2

SOH CAH TOA

3. The attempt at a solution

29a)

since we are given a 90 degree angle and the rest is 35 degrees, we can add
90+35

=125°

since we now have 2 sides of a triangle and the middle angle, we can use cosine to find the net force.

c^2=a^2+b^2-2abCosC

=(38)^2+(22)^2-2(38)(22)Cos125

=1444+484-1672Cos125

=1928-1672Cos125

=1928-959

c^2=√969

c=31.12 N

Fnet=31.12 N

v This 29b) is the important one, because this is where im not sure which solution is the right one.

29b)
Since one side of a line has 180 degrees, and we are given 45 degrees, one can assume 180-45 degrees would give you the angle on the left side of Z.

I attempted 29b) in 2 different ways, im not sure which one is the better way and gives the correct solution.

In method 1: i deal only with the downward vector (10 N) and the north west vector (17N)
and since 180-45=135 we have the angle in the middle, which is 135 degrees.

so using this information i use the cosine law to find the net force.

c^2=a^2+b^2-2abCosC

=(17)^2+(10)^2-2(17)(10)cos135

=289+100-340cos135

=389-240.4

=√148.6

Fnet=12.19 N

Method 2: This method, looks at the entire diagram, and assumes its one big right angle triangle.

using the smaller right triangle, we find side x.

Sin45=opposite/hypotenuse

sin45=x/75

sin45(17)=x

12.02N = x

(other side we use the given, which is 8 Newtons + 10 Newtons = 18 Newtons)

now that we have every side besides the hypotenuse we use Pythagorean to find the hypotnuse which will give us the net force.

c=√a^2+b^2

=(18)^2+(12)^2

=√324+144

=√468

c=21.63N

Fnet = 21.63 N

29c)

here we add the 2 given angles.

resulting in the middle angle, which is 32+24=56 °

since we have the 2 sides and the middle angle, once again we use cosine law, to find net force

c^2=a^2+b^2-2abCosC

=(15)^2 + (12)^2 -2(15)(12)Cos56

=225+144-360cos56

=√369-201.3

=√167.7

=12.94 N

Fnet= 12.94 N

File size:
14.5 KB
Views:
186
File size:
11.4 KB
Views:
169
File size:
13.3 KB
Views:
139
2. Mar 2, 2012

### SammyS

Staff Emeritus
Re: Net force (Diagrams incl) Please check my solutions :)

What you are doing is vector subtraction, not vector addition.

3. Mar 2, 2012

### supernova1203

Re: Net force (Diagrams incl) Please check my solutions :)

um..what?

4. Mar 3, 2012

### SammyS

Staff Emeritus
Re: Net force (Diagrams incl) Please check my solutions :)

I only looked at your figures. The resultants you show in your figures are consistent with subtracting one of the vectors, in each part, rather than adding all the vectors for each part.

I appears that the numerical results you have for the magnitudes of the resultant forces is correct. If you also need the direction of the resultants, your figures will lead you to the wrong answer in each part.

5. Mar 3, 2012

### supernova1203

Re: Net force (Diagrams incl) Please check my solutions :)

oh i have the directions, i just didnt post them because there were LOTS of calculations involved, and then id have posted a wall of text, i wanted to minimize the amount of work required by the good people of the forums to check my solutions, in any case 29b) i have 2 possible solutions, which one do you think works?

6. Mar 4, 2012

### supernova1203

Re: Net force (Diagrams incl) Please check my solutions :)

anyone? for 29b) ?

7. Mar 4, 2012

### Staff: Mentor

Re: Net force (Diagrams incl) Please check my solutions :)

Vectors are added graphically by placing them tail to tip in sequence. The resultant is the vector from the origin to the tip of the final vector in the chain.

File size:
4.4 KB
Views:
374
8. Mar 4, 2012

### supernova1203

Re: Net force (Diagrams incl) Please check my solutions :)

hm..so approach 1 was correct and the resultant net force is 12.19N? for 29b?

9. Mar 4, 2012

### Staff: Mentor

Re: Net force (Diagrams incl) Please check my solutions :)

The result does not look right.

To mathematically do the equivalent of the graphical method, find the individual components of all the vectors (E-W components and N-S) components, corresponding to X and Y components on a Cartesian X-Y plot) and add the like components. Can you write the components of each vector? Make a vertical list of the x-components and another of the y-components. Add the columns to find the components of your resultant.

10. Mar 5, 2012

### supernova1203

Re: Net force (Diagrams incl) Please check my solutions :)

hm... we havent learned how to add/subtract vectors in our course yet...maybe we will later on (there are still 3 sections left to this unit)

Anyhow, is there any way you can show me how to do this, in plain english please? :)

11. Mar 5, 2012

### Staff: Mentor

Re: Net force (Diagrams incl) Please check my solutions :)

A picture is worth a thousand words Here's an example showing two vectors A and B being added. Each vector is specified by its magnitude (length) and angle to a reference direction, usually the positive X-axis. Note that vector B has its tail end placed at the tip end of vector A so that whatever quantities are represented by A and B applied successively. In other words, they are summed.

The resultant vector is the 'result' of adding A and B. The component of the resultant along the X-axis is equal to the sum of the X-components of A and B. Similarly, the Y-axis component of the resultant is equal to the sum of the Y-components of A and B. Use whatever trigonometry methods you know to extract the individual components.

$R_x = A_x + B_x$

$R_y = A_y + B_y$

$R = \sqrt{R_x^2 + R_x^2}$

$\theta_R = atan \left(\frac{R_y}{R_x}\right)~~~$ note: beware of quadrant placement when dealing with negative valued components!

#### Attached Files:

• ###### Fig1.gif
File size:
68.8 KB
Views:
361
12. Mar 6, 2012

### supernova1203

Re: Net force (Diagrams incl) Please check my solutions :)

hm... thanks a lot!

So what your saying is each point on the resultant is the sum of a point on vector A and B

so the x coordinate of a point on resultant is the sum of an x coordinate on vector A and B?
and likewise for each y coordinate on the resultant?

Also if possible can you give me an example of subtracting vectors? Similarly to how you showed me how to add vectors in the above post :)

Or maybe to subtract vectors would be essentially the same thing as adding except, in negative axis? (x and y)

13. Mar 6, 2012

### Staff: Mentor

Re: Net force (Diagrams incl) Please check my solutions :)

The only point on the resultant that you're ever likely to be interested in is the point at the tip (where the arrowhead is). The distance from the origin to that point is the magnitude of the resultant vector, and the angle that a line segment drawn from the origin to that point makes an angle with the positive x-axis that defines the direction of the resultant vector.

Vectors are quantities with magnitude and direction. Vectors are not line segments, although they are represented that way as a handy graphical notation which facilitates thinking about the math.
Why don't you make an attempt at drawing the vector subtraction operation yourself? You should be able to find plenty of examples of via a "Vector subtraction diagram" web search.

14. Mar 8, 2012

### supernova1203

Re: Net force (Diagrams incl) Please check my solutions :)

hm.. ok i think i finally figured out this vector addition/subtraction thing

i did the diagram again, and got it done, my question is to find resultant force(mathematically) ill have to use cosine, since i cant use Pythagorean.

My question is, the original 45 degree angle, its between the resultant vector and the 17 N vector right?

So i should be able to use it in calculations via cosine to find resultant?

#### Attached Files:

• ###### PF question final 1.jpg
File size:
7.5 KB
Views:
106
15. Mar 8, 2012

### Staff: Mentor

Re: Net force (Diagrams incl) Please check my solutions :)

It is given that the 17 N vector makes an angle of 45° with respect to the +y axis (and because it's 45°, by symmetry it's also 45° from the -x axis). There's no particular "given" angle between a vector and some resultant that is the result of summing that vvector with a bunch of other vectors.

The procedure for summing vectors is to take each of the vectors individually and establish their x and y components. This is done using trigonometry and any angles you've been given that establish the vector's direction, and hence it's projections along the x and y axes of the coordinate system. The x components for each vector are summed to given the x component of the resultant. The y components for each vector are summed to given the y component of the resultant. These components define the result vector.

16. Mar 9, 2012

### supernova1203

Re: Net force (Diagrams incl) Please check my solutions :)

for the very first one (29a) since i know now that vectors are always added head/tail method, is the diagram i did incorrect? shouldnt it be 38 N vector, with the 22N vectors tail on the 38 N vectors head?

17. Mar 9, 2012

### Staff: Mentor

Re: Net force (Diagrams incl) Please check my solutions :)

Yes.
Yes, or the 22N vector with the 38N vector tacked onto its head. The order does not matter.