Calculate the rope tensions (please tell me if I've made a mistake)

[ScullyX51]

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.

 

[alphysicist]

Hi ScullyX51,

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.

I don't see anything wrong, and your final answer looks right to me.

 

[ScullyX51]

ok thanks. now if the problem asks what would happen if the box was moving up at constant acceleration, how would the equation change. I figured that since it is moving up there is only motion in the j direction, and so I substituted a into the equation for the j's and got:
t2sin(theta)+t2sin(theta)-mg=ma
2t2sin(theta)=mg+ma
t2= mg+ma/2sin(theta)
and since t1 is still equal to t2, this is the tension of t1 as well.
how does this look?

 

[alphysicist]

ok thanks. now if the problem asks what would happen if the box was moving up at constant acceleration, how would the equation change. I figured that since it is moving up there is only motion in the j direction, and so I substituted a into the equation for the j's and got:
t2sin(theta)+t2sin(theta)-mg=ma
2t2sin(theta)=mg+ma
t2= mg+ma/2sin(theta)
and since t1 is still equal to t2, this is the tension of t1 as well.
how does this look?

If I'm visualizing the problem correctly, this looks right to me, as long as when you type in:

t2= mg+ma/2sin(theta)

you mean

t2= (mg+ma) / 2sin(theta)