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## Homework Statement

Two students want to raise a heavy box onto the back of a truck. They do this by putting a rope through a handle on the box, and each lifting one side of the rope. The arrangement is symmetrical, so that the rope makes an angle (theta) with the horizontal on either side. The box is moved at constant speed.

1) draw a FBD

2) choose a coordinate system and write each of the forces in terms of our coordinates using vector notation.

3)Calculate the magnitude of the tension in each side of the rope. Let the mass of the trunk (including the handles) be M, and neglect the mass of the rope. In terms of M, g, and theta

## Homework Equations

F=ma

constant speed= 0 acceleration.

## The Attempt at a Solution

For the FBD I set up a standard xy coordinate system. I have mg pointing down in the j direction (or -y). T1 point diagonaly up to the left, and T2 pointing diagonaly up to the right. and M at the origin of course.

For the coordinates of the forces I have:

Fgrav= -mgJ

T1= t1(cos(theta)i+sin(theta)j)

T2=t2(-cos(theta)i+sin(theta)j)

To solve for the magnitude:

I gathered terms in the i direction:

T1cos(theta)-T2(cos(theta)=0

T1cos(theta)=T2(cos(theta)

T1= T2(cos(theta))/cos(theta)

the cos(theta) cancel out and we are left with: T1=T2

I gathered terms in the j direction:

T1sin(theta)+T2sin(theta)-mg=0

I then plugged in T2 FOR T1, since I calculated that from the I direction, so now I have:

T2sin(theta)+T2sin(theta)+mg -----> 2T2sin(theta)-mg------> T2= mg/2sin(theta) and since T1=T2 this is the answer for both sides of the rope.

I know this is a lot of text, but if someone could tell me if I did this right it would be greatly appreciated.