Calculating Force: Tractor on Trailers on Incline

[taveuni]

I am finding myself stuck on what is likely considered a very basic problem. I know the answer (it's in the book) but after 20 minutes of searching and trying to find a method, I've turned to you. Could someone please describe the methodology for this?

A farm tractor tows a 4300kg trailer up a 26-deg incline at a steady speed of 3.0m/s. What force does the tractor exert on the trailer? (ignore friction).

*This is a section in the book before Normal and frictional forces, so I assume they are not incorporated
*If the velocity is constant, isn't the acceleration then zero, thus making F zero?
*Is there some equation that I am missing that links mass, Fx=Fcos26, Fy=Fsin26, and speed?

 

[Pyrrhus]

Hello, Welcome to PhysicsForums!.

This problem is solved by Newton's 1st Law (in vectorial form):

$$\sum_{i=1}^{n} \vec{F}_{i} = \vec{0} \rightarrow \vec{v} = constant$$

*If the velocity is constant, isn't the acceleration then zero, thus making F zero?

Not true, it only makes the sum of forces equal to 0, as the 1st Law stated above says.

In order to solve this problem you will need to notice the forces acting on the cargo are the gravity force, the normal force and the force applied by the tractor to the cargo. I suggest putting your coordinate system at the angle of the slope, and using standard sign convention.

 

[taveuni]

Right, I understand what you are saying. However, this is a part in the book BEFORE any mention of weight or normal forces. So they cannot be incorporated into solving the problem. So I feel like the only force at play here (given that fact) is the force of the farmer. Right?

A similar problem is about a 65kg skier who speeds down a smooth trail at an angle of 22-deg. They ask for the direction and magnitude of the force. How do you get that? I am stuck at the simplest of equations.

 

[Pyrrhus]

Right, I understand what you are saying. However, this is a part in the book BEFORE any mention of weight or normal forces. So they cannot be incorporated into solving the problem. So I feel like the only force at play here (given that fact) is the force of the farmer. Right?

Not exactly. In fact you don't need to normal force, to solve this problem. Simply consider a sum of forces along the x-axis rotated at 26 degrees, you will only have the x component of the gravity force or weight and the force applied by the tractor which is tangential to the slope.

A similar problem is about a 65kg skier who speeds down a smooth trail at an angle of 22-deg. They ask for the direction and magnitude of the force. How do you get that? I am stuck at the simplest of equations.

Can you be more explicit?, the problem doesn't make sense without the complete statement.

 

[taveuni]

I will look into the tractor problem. Would the gravity force incorporate weight (mg)?

As for the other problem (these are from the book - I am "practicing", albeit unsuccessfully): "A 65-kg skier speeds down a trail. The surface is smooth and inclined at an angle of 22 deg with the horizontal. Find the direction and magnitude of the net force acting on the skier." They somehow obtain an answer of 240N from that information. But how, conceptually?

 

[Pyrrhus]

I will look into the tractor problem. Would the gravity force incorporate weight (mg)?

When i said gravity force i meant the weight.

As for the other problem (these are from the book - I am "practicing", albeit unsuccessfully): "A 65-kg skier speeds down a trail. The surface is smooth and inclined at an angle of 22 deg with the horizontal. Find the direction and magnitude of the net force acting on the skier." They somehow obtain an answer of 240N from that information. But how, conceptually?

There's a non-null resultant force in this case. If you notice the forces acting on our body(the skier) don't add vectorially to zero, ergo our body is accelerating.

In general, to solve most of this problems, you must isolate the object of interest, and draw all the forces acting on it, and then apply Newton's Laws accordingly. This method is called the Free Body Diagram or FBD.