Tension in rope supporting box over pulley with two rope angles

[tony873004]

A box of mass M = 5.7 kg is suspended by a rope over a pulley (Figure 5.64). Find the tension in the rope supporting the box and in the rope supporting the pulley.

The tension in the rope supporting the box should just be mg? right?

But the two labeled angles are not the same. Combined, they must be equal to -mg or the pulley would accelerate. But without knowing theta angle, I'm not sure how to do this? Do they want the tension of the rope supporting the pulley to be expressed in terms of something else, or is it possible to come up with a precise number in Newtons?

 

[Doc Al]

The tension in the rope supporting the box should just be mg? right?

Right.

But the two labeled angles are not the same. Combined, they must be equal to -mg or the pulley would accelerate. But without knowing theta angle, I'm not sure how to do this?

Hint: the tension in the rope wrapped around the pulley is the same throughout.

 

[tony873004]

Hi, DocAl. Thanks for your reply. I got an answer. Can you tell me if I did it correctly.

It would seem to me that tension in the rope attached to the box would just be mg.

So the rope supporting the weight would have a tension of mg in the y direction and 0 in the x direction.

The rope after it passes through the pulley also has a tension of mg which can be broken up into its components by fx = mg cos 37, and fy = mg sin 37.

Adding this to the components of the part of the rope holding the box, I get

fx = cos37 * 5.7 * 9.9
fy = sin37 * 5.7*9.8 + 5.7 * 9.8

which should be the resultant force (tension in the other rope) needed to make sure the pulley does not accelerate. Plugging it into the calculator:

fx = 44.61 N
fy = 89.48 N

Using pothag, tension in the other rope is sqr (44.61² + 89.48²) = 99.98 N.

So the final answer:
tension in the rope supporting the box = 55.86 N away from the box
tension in the rope supporting the pulley = 99.93 N away from the pulley

Did I do it right??

And tension is a force, therefore a vector. Is it correct to give it direction. For example, the one attached to the box pulls in both directions. It pulls the box up and it pulls the pulley down. Or should I express tension as a scalar and not worry about direction?

 

[Doc Al]

So the final answer:
tension in the rope supporting the box = 55.86 N away from the box
tension in the rope supporting the pulley = 99.93 N away from the pulley

Did I do it right??

Looks good to me. (See comments below about describing tension.)

And tension is a force, therefore a vector. Is it correct to give it direction. For example, the one attached to the box pulls in both directions. It pulls the box up and it pulls the pulley down. Or should I express tension as a scalar and not worry about direction?

The force exerted by a rope (due to the tension in the rope) is of course a vector. The direction of the force is along the line of the rope (and ropes can only pull). But the tension is a property of the rope, best described as a scalar. (As long as there is no danger of confusion, it's standard to refer to the force exerted by the rope as "the tension", but that's not strictly accurate.)

 

[Nenad]

Just wait until you get into pulles that have mass. You thought this was fun. It gets more fun.

Regards,

Nenad

 

[arildno]

Just wait until you get into pulles that have mass. You thought this was fun. It gets more fun.

Regards,

Nenad

It gets even more fun when the ropes are given mass..