Ok, I wont pretend to know anything about cars, because I dont know much about them. But I am studying engineering (aerospace) and know a bit (a small bit) about how gears work. Also feel free to correct me if its wrong, I only just finished first year and have only done 1 unit which included gears.
Lets work in metric so my head doesn't explode and I'll try and keep it simple. The easiest way I can think of doing it is in terms of power, speeds and torques, you can do forces and gear ratios if you want and I hope you'd come to the exact same answers.
Lets say car is moving 20m/s, which is a bit more than 60kmh and the resistive forces at the wheels are 4000N (doesn't matter whether the car is accelerating or not). This means the power which needs to be supplied to the wheels is 20x4000=80000=80kW.
This is what the engine needs to supply to the wheels, and we'll ignore any power losses (of which there are many).
If you're in a gear where the engine is revving, say, 500rad/s (a bit under 5000rpm, so probably 1st or 2nd gear), and the engine then has to be supplying 80kW, so it must be delivering 80000/500 = 160Nm of torque.
Now lets say the engine is going 100rad/s, probably 5th or 6th gear, the engine now still needs to supply 80kW, but the torque it needs to supply is 80000/100 = 800Nm.
This means the clutch, which is connected to the engine, has to be transferring 5 times as much force. Or if the clutch is engaging/disengaging it has a change in force which is much greater (though this change in force depends on the speed and momentum of the car and engine, clutch, gearbox, etc when the clutch is engaged/disengaged and isn't hugely simple to calculate).
Now I didn't actually use any gear ratios, simply the difference in angular velocity of the engine, which is obtained by having different gear ratios, if you knew the diameter of the wheel you could calculate the gear ratios from it and you'd get the same result (I hope, unless I've made some fatally stupid mistake, which I often do especially given its absurdly late at night, so feel free to correct me).
Conclusion: Driving in 6th gear at low speed places larger torques on the engine, clutch and gearbox.
EDIT: Actually I'm an idiot... it is easier to explain in terms of gear ratios, I'm just not thinking straight.
Imagine the torque at the rear wheels is, say, 1600Nm (due to friction, wind resistance, acceleration, etc). Lets say you're in first gear with a gear ratio of 10:1... that means for every 10 revolutions of the engine, the rear wheels revolve once. This means the torque applied back through the gearbox changes from 1600Nm down to 160Nm. Now imagine the gear ratio is 2:1 (more than likely an overdrive 6th gear). The torque at the rear wheels is still 1600Nm (the car has the same resistances and same acceleration), but after gearing down the torque at the engine is now 800Nm.
Gezmoor, you're right the forces at the clutch have to be balanced otherwise the clutch is slipping. What you're neglecting is the fact the gearbox changes the amount of torque delivered from the wheels before it ever reaches the clutch. You can think of it in a few different ways. One that power = force (or torque) x speed (or rotational speed), if the speed decreases force must increase in a system where power (and thus energy) is not lost. Another way is to consider the forces at the contact patches where the gears mesh, these also have to be balanced so the gears dont slip, for 2 gears of different sizes (number of teeth) for the force at the contact patch to be balanced, the torques on the gears themselves must be different (torque = force x distance from rotational center).