high school physics force question ......can someone help?

Question

hi .... i'm hoping someone can help me with a problem from my high school physics book ..... i've worked on this problem for about 1 hour and can't seem to get the answer ...... my teach didn't ask us to do this one as homework but i'm trying it any way .....i'm stuck ....... i hope someone can help..... thanks.....

Three coplanar forces act on a 75000.0 kg object in outer space, where there is no gravity: 14.2 kN directed at 357°, 25.6 kN at 138°, and 16.4 kN at 255°. All directions are with respect to the x direction.

What is the magnitude and direction of the acceleration?

If the object has an initial velocity of 17.5 miles/h directed at 58.6° in the same plane as all three forces, what will its velocity and position be after the force has been applied for 65.5 s?

What will the objects direction from the origin be and in what direction will it be moving assuming the force was applied at the origin of coordinates?

Answer 1

Other answer is good. Forces can be reduced to their components along the coordinate axes, then added. The final two components make up the two sides (for 2 dimensions) of the triangle.
So 1) reduce all the forces to their x & y components
(best to draw a rough diagram first to make sure you're getting the signs right)
2) add the x's together
3) add the y's together
the final force is in the direction of the angle A where X/Y=tan(A)
(or is it Y/X .. i've forgotten!!!) and its magnitude is Sqrt(X^2 +Y^2)
- thats the hypotenuse of the triangle, don't you know.
Tricky part is to keep all the signs of the components correct.
Pick your origin and x&y axes and if (in your rough drawing) the force is below the axis the y component better be negative, if the force is to the left (of the origin) then the x comp. is negative, right?
Sin(a)=opp/hyp, Cos(a)=adj/hyp
Must use calculator to figure arctan(X/Y)=A (or tangent tables)
You'll have to find out if tan is opp/adj (X/Y) or vice versa, sorry.

To the second part. I guess you know the formulas for distance and velocity which includes acceleration? d=d0+v0*t+1/2*a*t^2
and v=v0+a*t, right? ALL you have to do is know that both of these work with components, too. So since you already know the force ( X & Y) and since F=ma you can do all four equations by adding the components. dy (distance in y direction) = dy0 +vy * t + 1/2*a*t^2 so you know t , you figured ay=Fy/m
and you can figure the y component of the given velocity.
Figure the vx and ax the same way.
Solve for dy, dx, vy,vx and do the two triangels for direction and magnitude (the magnitude of v is the speed, of the d is the distance) like above.

Answer 2

this is a question about vector addition.

the easiest way to add vectors is to add thier x andy components.
then you add them up and use (X^2+y^2)^(1/2) to get the total maginitud. This will be your force vector. For example, the x and y components of 14.2kN at 357 degrees = cos(357)14.2 kN and sine(357)*14.2 kN

note it is very important in this problem to draw a good free body diagram and to keep trakck of positive and negative numbers

the rest of the problem is pretty simple once you get part a finished

Answer 3

You don't need to solve for all three forces at once. Just solve for two and treat that result as a single force and solve it again with the third.

Answer 4

via fact the two angles are 30°, the ceiling has the comparable vertical comprehensive via fact the 16Kg mass. The mass M provides the horizontal matching tension. So the horizontal is: Fx = Mg = 2*[ 16g cot(30°)] M = 2*[ sixteen cot(30°)] = fifty 5.4 Kg