Another motion question

[Inside Out]

1.2 7)b. During preseason football training, Matt was required to run with a bag of sand dragging behind him. The bag of mass 50kg was attached to a rope, which made an angle of 25 degrees to the horizontal. When Matt ran with a constant speed of 4.0 m/s a frictional force of 60N was acting on the bag.    Calculate the size of the tension force acting in the rope. Answer:66N         The net force is zero because of the constant velocity, so all the forces acting on the bag have to balanced. Does this include the force of the rope pulling the bag. Why do the solutions only consider the frictional and horizontal force having to be balanced?   Please help

[Aurelian]

1.2 7)b. During preseason football training, Matt was required to run with a bag of sand dragging behind him. The bag of mass 50kg was attached to a rope, which made an angle of 25 degrees to the horizontal. When Matt ran with a constant speed of 4.0 m/s a frictional force of 60N was acting on the bag.    Calculate the size of the tension force acting in the rope. Answer:66N         The net force is zero because of the constant velocity, so all the forces acting on the bag have to balanced. Does this include the force of the rope pulling the bag. Why do the solutions only consider the frictional and horizontal force having to be balanced?   Please help

The frictional force runs opposite to movement. Movement is in the horizontal direction, and therefore so is the frictional force is as well, but in the opposite direction.

The bag moves at a constant velocity, and so the net force on the bag must be zero, as said in the answers. Thus, the horizontal component of the tension (ie the 'horizontal force') must equal the frictional force to cancel out. That is, it must also be 60N.

Now we use trigonometry. I'd draw a triangle but... I can't

But if you draw it yourself, with the angle of 25 degrees and the horizontal being 60N, you simply use cosine (adjacent over hypotenuse);
cos(25) = 60/hyp
hyp = 60/cos(25) = 66N

The hypotenuse represents the actual rope.

Hope that helps!

[Inside Out]

What about the force from the person pulling on the rope. Wouldnt it do something to 'knock out' the constantvelocity/ zero net force of the bag?

[Lasercookie]

What about the force from the person pulling on the rope. Wouldnt it do something to 'knock out' the constantvelocity/ zero net force of the bag?

The force from the person pulling on the rope is the reason why the bag moves and is also the reason why it has a zero net force. The person is moving at a constant velocity, which means the bag also moves at a constant velocity.

Without that force from the person, the bag wouldn't be moving.

[Aurelian]

What about the force from the person pulling on the rope. Wouldnt it do something to 'knock out' the constantvelocity/ zero net force of the bag?

You need to view the force from the person pulling on the rope as actually two component forces (this is what we do in the triangle). One component is the horizontal force (what we calculated out) and this is cancelled by friction. The other component is the vertical force (that is, the part of you pulling 'up' on the rope); this component is cancelled out by gravity.

[Inside Out]

So is the horizontal and the friction force a action-reaction force?

[Inside Out]

Pair*

[Aurelian]

So is the horizontal and the friction force a action-reaction force?

Ehhh no not really... The horizontal force isn't really a force in and of itself, it's just a component of the force along the rope. The action-reaction pair would be "you pull the rope and the rope pulls you equally and oppositely".