Resultant Force of Two Applied Forces on a Car

[sparky450r]

Two forces, 489 N at 11
◦
and 444 N at 25
◦
are
applied to a car in an effort to accelerate it What is the magnitude of the resultant of
these two forces?
Answer in units of N.

 

[tiny-tim]

hi sparky450r! welcome to pf!

components should do it …

show us what you've tried, and where you're stuck, and then we'll know how to help! ​

 

[sparky450r]

Problem shouldn't be that hard. I have done examples identical to this and got them correct but the computer won't accept my answer for this one!

First I summed X and Y

x=356cos(9) + 313sin(26)

and y
y=356sin(9) + 313cos26

Then I squared these two values, added them, and took the square root.

all wrong!

 

[willem2]

why do you use sin for the first term and cos for the second term in the equaton for x?
you seem to use different number than in the first image.

make a drawing of the forces and the components, so you can see in which direction they point and when you have to add and when to subtract them

 

[sparky450r]

why do you use sin for the first term and cos for the second term in the equaton for x?
you seem to use different number than in the first image.

make a drawing of the forces and the components, so you can see in which direction they point and when you have to add and when to subtract them

Yea sorry I was doing a different version of the problem in my last post.

So assuming we are using the first picture...how would I know when to add and subtract and which go in the X equation and the Y because I am obviously missing something here.

 

[tiny-tim]

hi sparky450r!

x=356cos(9) + 313sin(26)

and y
y=356sin(9) + 313cos26

why do you use sin for the first term and cos for the second term in the equaton for x?

...how would I know when to add and subtract and which go in the X equation and the Y because I am obviously missing something here.

as willem2 points out, your cos and sin are getting confused

the rule is …

it is always always ALWAYS cos, of the angle between the force and the direction of the component …​

the only time it looks like sin is when the angle you're given is the "wrong" angle, ie 90° minus the correct angle …

in that case, it's still cos, but it's cos(90° - the angle), ie sin

so in this case, are the named angles of 11° and 25° the "correct" angles, or not? if they are, then use cos ​

 

[sparky450r]

Are you saying use cos to compute both x's and both y's. no sin?

 

[sparky450r]

Are you saying use cos to compute both x's and both y's. no sin?

And yes they are correct angles.

 

[tiny-tim]

Are you saying use cos to compute both x's and both y's. no sin?

for the x direction, use cos of the angle between the two forces and the x direction

for the y direction, use cos of the angle between the two forces and the y direction

 

[sparky450r]

Ok, finally got it.

Y=356sin9-313sin26
X=356cos9+313cos26

These values are for the latter problem that I posted. Not the first one.

 

[tiny-tim]

Y=356sin9-313sin26
X=356cos9+313cos26

yes that's it!

(and then of course the squarey thing to find the magnitude )