- #1

Jamotron

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## Homework Statement

Two ropes are attached to a 40kg object. The first rope applies a force of 45 N, and the seond 40 N. If the two ropes are perpendicular to each other, what is the resultant acceleration?

a)1.2m/s

^{2}

b)3.0m/s

^{2}

c)25m/s

^{2}

d)47m/s

^{2}

## Homework Equations

F=ma

F

_{1x}=F

_{1}cos theta

_{1}

F

_{2x}=F

_{2}cos theta

_{2}

F

_{1y}=F

_{1}sin theta

_{1}

F

_{2y}=F

_{2}sin theta

_{2}

a=(a

_{x}

^{2}+a

_{y}

^{2})

^{1/2}

## The Attempt at a Solution

As the angles of ropes to the object are not given but the ropes are stated as being perpendicular I took 45 degrees as the angle of both ropes, which gives me the same answer for F

_{x}and F

_{y}which is49.45N. Therefore:

a

_{x}=F

_{x}/M = 1.23m/s

^{2}

a

_{y}=F

_{y}/M=1.23m/s

^{2}

Now, this value is consistent with answer a) given in the question. However I was under the impression that to find the magnitude of the acceleration I need to use the equation

a=(a

_{x}

^{2}+a

_{y}

^{2})

^{1/2}. As I've figured the x and y acceleration to be equal this will give me a =1.74 m/s

^{2}.

So where have I gone wrong? What would be the correct procedure for solving this problem? I'm very new to physics so any help would be appreciated :)