What is the tension in the coupling between two trucks on an inclined slope?

[xllx]

Homework Statement

A truck of mass 5kg is pulling a smaller truck of mass 2kg up a slope inclined at 10degrees to the horizontal. The trucks are joined together by a light, rigid coupling. The 5kg truck has a resistance of 8N and the smaller truck has a resistance of 6N. The string pulling the trucks is parallel to the horizontal and has a tension of 43N. The two trucks are accerelerating at 2.5ms-2. What is the tension in the coupling?

Also, show that the coupling remains in tension whatever the tension in the string is.

Homework Equations

F=ma

The Attempt at a Solution

I've drawn a diagram but I am struggling most on which way the tension in the coupling would be so whether to take it away or add it in the Force part of the equation. I've only done the equation for the first truck, but not sure whether it is right:

F=ma
Tension in string- Resistance- Weight component-Tension in coupling= ma
43-8-5(9.8)sin10-T=5x2.5
26.5-T=12.5
T= 14N

I really have no idea for the second part.

Any help at all would be greatly appreciated.

 

[LowlyPion]

Homework Statement

A truck of mass 5kg is pulling a smaller truck of mass 2kg up a slope inclined at 10degrees to the horizontal. The trucks are joined together by a light, rigid coupling. The 5kg truck has a resistance of 8N and the smaller truck has a resistance of 6N. The string pulling the trucks is parallel to the horizontal and has a tension of 43N. The two trucks are accerelerating at 2.5ms-2. What is the tension in the coupling?

Also, show that the coupling remains in tension whatever the tension in the string is.

Homework Equations

F=ma

The Attempt at a Solution

I've drawn a diagram but I am struggling most on which way the tension in the coupling would be so whether to take it away or add it in the Force part of the equation. I've only done the equation for the first truck, but not sure whether it is right:

F=ma
Tension in string- Resistance- Weight component-Tension in coupling= ma
43-8-5(9.8)sin10-T=5x2.5
26.5-T=12.5
T= 14N

I really have no idea for the second part.

Any help at all would be greatly appreciated.

Since all they want is tension in the connection to the second truck, then just consider the second truck in isolation.

You have a resistance of 6N, you have a component of gravity m*g*sin10 and you have the additional acceleration that is 2.5m/s² of the system times its mass.

The sum of those then should be the tension in the coupling between the lead truck that is being pulled with the string and the smaller truck.

 

[xllx]

Thankyou

So would this be right:

F=ma
T-6-2gsin10=2x2.5
T-25.3=5
T=30.3N

Thankyou again.

From that how would I show that the coupling remains in tension no matter what the tension is in the string?

 

[LowlyPion]

Thankyou

So would this be right:

F=ma
T-6-2gsin10=2x2.5
T-25.3=5
T=30.3N

Thankyou again.

From that how would I show that the coupling remains in tension no matter what the tension is in the string?

I would want to check your value for sin10 degrees. (I get .174)

For part 2) consider the case where there is no tension in the string, and the trucks are not accelerating.

 

[xllx]

I would want to check your value for sin10 degrees. (I get .174)
QUOTE]

Sorry, miscalculation. So it would be:
T-9.4=5
T=14.4N

I'm still slightly confused on the second part. When there's no tension in the string, that would mean that it would not be moving up the slope, so the coupling would be in compression and so if a=0, then
F=ma
6+8+2gsin10+5gsin10+T=0
T=-25.9N

Is this anywhere remotely right?

 

[LowlyPion]

I'm still slightly confused on the second part. When there's no tension in the string, that would mean that it would not be moving up the slope, so the coupling would be in compression and so if a=0, then
F=ma
6+8+2gsin10+5gsin10+T=0
T=-25.9N

Is this anywhere remotely right?

Is there a picture with this problem?

 

[xllx]

I've attached a picture, sorry had to draw it on paint.

If the tension in the coupling comes out as a negative does that mean that it is in compresion or just wrong?

thanks