Two forces are applied to a car in an effort to accelerate it, as shown below. The first force, F1 = 330 N, is applied at an angle α = 38° to the forward dashed line. The second force, F2 = 532 N, is applied at an angle β = 14° to the forward dashed line.

http://www.webassign.net/holtphys/p4-22alt.gif

(a) What is the resultant of these two forces?
N at ° to the

of the forward dashed line

(b) If the car has a mass of 2900 kg, what acceleration does it have? (Disregard friction.)
m/s2 at ° to the

of the forward dashed line

Question

Two forces are applied to a car in an effort to accelerate it, as shown below. The first force, F1 = 330 N, is applied at an angle α = 38° to the forward dashed line. The second force, F2 = 532 N, is applied at an angle β = 14° to the forward dashed line.

http://www.webassign.net/holtphys/p4-22alt.gif

(a) What is the resultant of these two forces?
  N at   ° to the  

 of the forward dashed line

(b) If the car has a mass of 2900 kg, what acceleration does it have? (Disregard friction.)
  m/s2 at   ° to the  

 of the forward dashed line

anonymous · Answer

common! muntooo!!!

anonymous · Answer

Haha. OK:

First, you have to sum the forces. Do you remember how to do that?

anonymous · Answer

muntoo its been a long night its 12;11 can u just give me th formula with the numbers plugged in

anonymous · Answer

I'm actually a grade level lower than this, so I don't know what the formula is. However:

$$\Large {F_x = F_1 \cos (\pi/2 -  \alpha) + F_2 \cos (\pi/2 + \beta) \ F_y = F_1 \sin (\pi/2 -  \alpha) + F_2 \sin (\pi/2 + \beta) }$$
...LaTeX takes forever to type out...

Disclaimer: I might be completely wrong.

anonymous · Answer

Then,

$$\Large F = \sqrt{F_x^2 + F_y^2}$$

and the angle:

$$\Large \theta = \arctan ({F_y \over F_x})  - \pi/2   $$