Physics question, I can't seem to get this problem, please help.?

Question

Two forces, 381N at 8deg. & 307N at 25deg. are applied to a car in an effort to accelerate it.?
What is the magnitude of the resultant of
these two forces?

What is the direction of the resultant force (in re-
lation to forward, with counterclockwise con-
sidered positive)?????
within the
limits of -180deg to 180deg

655.52 was the answer for the first question... and so if 655.52 is the resultant force, what is the direction of deg?

Answer 1

Remember that force is a vector, meaning that is has both magnitude and direction. In this case, the two vectors are:

A = 381 @ 8° and B = 307 @ 25°

The resulting force would be the addition of the two component forces. To do this we must look at the independant y and x components of each vector by using some trigonometry.

y component of A = 381 sin 8 = 53
x component of A = 381 cos 8 = 377

y component of B = 307 sin 25 = 130
x component of B = 307 sin 25 = 278

Adding the components together:

y component of resulting = 53+130= 183
x component of resulting = 377+278 = 655

The over-all magnitude of the resulting force is found using the Pathagorean Theorem.

y^2 + x^2 = c^2

c^2 = 462514

c = 680

For the second question, we need to use some trig again.

Tan (theta) = y / x

Theta = Arctan (y/x) = Arctan(183/655) = 15.6°

Answer 2

Break each piece into components

Fx1 = F1 cos theta1
Fx2 = F2 cos theta2
Fy1 = F1 sin theta1
Fy2 = F2 cos theta2

Add up like components

Fx = F1 cos theta1 + F2 cos theta2
Fy = F1 sin theta1 + F2 sin theta2

Use pythagorus to combine them
F = sqrt (Fx^2 + Fy^2)

Finally, the direction is given by
theta = arctangent (y/x)

Have fun.

Answer 3

get a calculator that can to vector stuff, like prettymuch any scientific calculator costing more then ten dollars:


(381@8) + (307@25) =(681@16)

the direction is 16 degres (15.5790684729 more presicely)