Friction between tractor tyres and the ground

[laser]

Homework Statement

See description

Relevant Equations

F=ma

So in (a) the answer is just mgsintheta.

When doing (b), I got 8.44x10^3N to be the NET force the tractor must exert. Does the tractor not need to overcome its own friction as well? If so, shouldn't the equation be F_tractor = F_exerted - F_friction, and the question is asking for F_exerted?

How I got 8.44x10^3: a = 0, therefore mgcostheta*Uk +mgsintheta = F_tractor. So the tractor is applying a net force of 8.44x10^3. But why are we not considering its friction? Consider my previous point.

Given that the coefficient of friction between the tractor's tyres and the ground is not given, can I assume it is frictionless? But then, how are the wheels meant to rotate? :)

 

[BvU]

So in (a) the answer is just mgsintheta.

Which, of course, you checked agains $\mu N$ ?

Does the tractor not need to overcome its own friction as well?

Yes

If so, shouldn't the equation be F_tractor = F_exerted - F_friction, and the question is asking for F_exerted?

Apparently the exercise composer considers F_tractor to be the net available traction force you call F_exerted
$\$

 

[kuruman]

Does the tractor not need to overcome its own friction as well?

What does that mean? As you say, the tractor would be unable to pull on the rope if it were on frictionless ice. It's the external force of static friction between the tyres and the ground that propels the (tractor + rope + logs) system forward.

Think of it this way. There is no friction to "overcome" before engaging the gears that transmit the engine power to the wheels ." This situation is unlike a fish swimming upstream that has to overcome a current. The only static friction that must be overcome is between the box of logs and the ground not between the tractor's tyres and the ground. Static friction adjusts itself to provide the observed acceleration but only up to a point that cannot be exceeded.

Here, as the tension in the rope increases while nothing moves the force of static friction at the box and the tyres match the tension. In order for the tractor to pull the box up the slope, the upper limit of static friction at the box must be less than the upper limit of static friction at the tyres. If it isn't, the tyres will start spinning before the box is pulled uphill. See how it works?

How I got 8.44x10^3: a = 0, therefore mgcostheta*Uk +mgsintheta = F_tractor. So the tractor is applying a net force of 8.44x10^3. But why are we not considering its friction? Consider my previous point.

Your cavalier use of "net force" might lead you into trouble. According to Newton's second law $F_{\text{net}}=ma.$ You cannot have a non-zero force when the acceleration is zero. What you have computed is the force exerted by the tractor on the rope a.k.a. the tension in the rope.

 

[PeroK]

Homework Statement: See description
Relevant Equations: F=ma

View attachment 336895
So in (a) the answer is just mgsintheta.

When doing (b), I got 8.44x10^3N to be the NET force the tractor must exert. Does the tractor not need to overcome its own friction as well? If so, shouldn't the equation be F_tractor = F_exerted - F_friction, and the question is asking for F_exerted?

How I got 8.44x10^3: a = 0, therefore mgcostheta*Uk +mgsintheta = F_tractor. So the tractor is applying a net force of 8.44x10^3. But why are we not considering its friction? Consider my previous point.

Given that the coefficient of friction between the tractor's tyres and the ground is not given, can I assume it is frictionless? But then, how are the wheels meant to rotate? :)

The static friction between the tractor's tyres and the ground are what allows it to move. It's an accelerating force, not a force of resistance. That's how wheels work. And, that's how walking works as well.

There may be some rolling resistance that is being neglected. But definitely not the friction that applies to a sliding object.

 

[Lnewqban]

How I got 8.44x10^3: a = 0, therefore mgcostheta*Uk +mgsintheta = F_tractor. So the tractor is applying a net force of 8.44x10^3. But why are we not considering its friction? Consider my previous point.

Following the same logic that led you to calculate a), the net force is exactly the force of static friction exerted on the tires by the ground after the tractor starts to pull: 8451 Newtons.