Finding angle of resultant force from two applied forces

[PhysicsPadawan]

Homework Statement

Two forces are applied to a car in an effort to move it, as shown in the figure below. (Let
F1 = 430 N (10 degrees)
and
F2 = 356 N. (30 degrees)
Assume up and to the right are in the positive directions.)
(a) What is the resultant vector of these two forces?
magnitude=?
° to the right of the forward direction=?
(b)If the car has a mass of 3,000 kg, what acceleration does it have? Ignore friction.

Homework Equations

(430sin(10)) + 356sin(30)=252.67x
(430cos(10)) + 356cos(30)=731.78y
a=(731.78)/(3000)

F=ma
angle = tan-1((731.78y)/(252.67x)) ??[/B]

The Attempt at a Solution

For part a I got 774.17N for the magnitude and it was correct
For part b I got 0.244 for the acceleration and it was correct
But I can not for the life of me solve for angle to the right of the forward direction
[/B]

 

[SteamKing]

Homework Statement

Two forces are applied to a car in an effort to move it, as shown in the figure below. (Let
F1 = 430 N (10 degrees)
and
F2 = 356 N. (30 degrees)
Assume up and to the right are in the positive directions.)
(a) What is the resultant vector of these two forces?
magnitude=?
° to the right of the forward direction=?
(b)If the car has a mass of 3,000 kg, what acceleration does it have? Ignore friction.

Homework Equations

(430sin(10)) + 356sin(30)=252.67x[/B]

This calculation is incorrect. If you look at the diagram, the x-components of each force act in opposite directions, so you can't simply add them together to find the resultant force in the x-direction.

Always indicate units with your calculations.

(430cos(10)) + 356cos(30)=731.78y
a=(731.78)/(3000)

F=ma
angle = tan-1((731.78y)/(252.67x)) ??

These calculations are OK.

The Attempt at a Solution

For part a I got 774.17N for the magnitude and it was correct
For part b I got 0.244 for the acceleration and it was correct
But I can not for the life of me solve for angle to the right of the forward direction[/B]

If you are still having trouble, make a small sketch of the resultant force vector and F1 and F2. You should be able to find the correct angle once you have calculated the correct x-component of the resultant force.

 

[PhysicsPadawan]

When I recalculated my x-component of the resultant force I got 103.33 N and an angle of 81.96 degrees.
F2x-F1x= Resultant x-component
(356 N sin(30)) - (430 N sin(10))= 103.33 N
tan-1=((731.78)/(103.33)) = 81.96
would 81.96 be the degrees to the right of the forward direction then that the question is asking for?
Thanks for the clarification and help!

 

[SteamKing]

When I recalculated my x-component of the resultant force I got 103.33 N and an angle of 81.96 degrees.
F2x-F1x= Resultant x-component
(356 N sin(30)) - (430 N sin(10))= 103.33 N
tan-1=((731.78)/(103.33)) = 81.96
would 81.96 be the degrees to the right of the forward direction then that the question is asking for?
Thanks for the clarification and help!

Did you draw the resultant and F1 and F2 to see if this calculation is correct?

I think you have calculated the wrong angle.

Does it seem reasonable that the resultant of F1 and F2 would act at almost a right angle to them?