Force and Newton law of Motion

  • Thread starter zeshkani
  • Start date
  • #1
29
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hi there i have a question in which i can't get the answer i will show my work on what i did, if anybody can help , that would help alot .

here is the question:

only two forces act on an object (mass= 3kg)


x=40N
a=60N is at 45 degrees above the horizon
y= ?
find the magnitude and direction (relative to the x-axis) of the acceleration of the object?

i can use (a=60N)sin(45)= 42.42degrees, but where do i go from here, or whats the next step, i did try squaring y+x under the square root, but thats not the magnitude, because the answer in teh back of the book is different.

any help is welcome thanks:smile:
 

Answers and Replies

  • #2
489
0
Ok first of all the 42.42 value wont be in degrees, it will be in newtons. Secondly, draw a diagram, it always helps.

In order to help you more i need to know which direction the x force is in. so draw up what you think is happeneing and then attach it.

In essence you are just resolving vector components of a force, and need to draw a few triangles and stuff to find the resultant
 
  • #3
andrevdh
Homework Helper
2,128
116
Newton's second law can be applied in component form:

[tex]\Sigma F_x = ma_x[/tex]

that is

[tex]a_x = \frac{\Sigma F_x}{m}[/tex]

so to get the acceleration in the x-direction you need to add the force components acting on the object in the x-direction and divide that by its mass.
 

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