Susans 12 Kg Baby Brother Paul Sits on a Mat

mawalker

I am working on the following problem.

Susan'south 13.0 kg infant blood brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled thirty degrees above the floor. The tension is a constant 31.0 North and the coefficient of friction is 0.190 . Utilize piece of work and free energy to find Paul'south speed after existence pulled ii.60 m.

And so far I accept fix up my forcefulness diagram with my x-axis along the ground. I take normal forcefulness pointing upwards. weight pointing down. tension to the left. and force susan to the right. I'm really not too sure where to go from hither though. Any thoughts?

Hootenanny

Your tension should non exist pointing left, information technology should exist inclined at 30 degrees to the horizontal. Your next step should be to find the normal reaction force.

mawalker

Your tension should not exist pointing left, it should exist inclined at 30 degrees to the horizontal. Your next step should be to detect the normal reaction force.

thank you hootenanny. i stock-still my force diagram and i calculated the normal strength to be 13kg * 9.8 ms = 127.4 N. Am I on the right runway?

Hootenanny

cheers hootenanny. i fixed my force diagram and i calculated the normal force to exist 13kg * 9.8 ms = 127.4 Northward. Am I on the right rails?

Not quite. Look at all the vertical forces acting, you should have 3;

Weight of the child
Normal Reaction Force
?

What's the third?

mawalker

the strength of friction acting along the incline

Hootenanny

the force of friction acting along the incline

Not quite, there is no incline, the friction is interim horizontally. In add-on to the two forces outline above, you will also have a component of the tension acting vertically (and a component acting horizontally). Can you see why?

mawalker

yeah i call back so, so would the force of tension vertically just be the 31 N * sin (30 degrees) = xv.5 N added to the 127.four and i go 142.ix North full force. Is this right?

OlderDan

yeah i think so, so would the forcefulness of tension vertically but be the 31 Northward * sin (30 degrees) = 15.5 Northward added to the 127.four and i go 142.ix N total force. Is this correct?

Accept a careful look at the directions of the vertical forces.

mawalker

right, the tension strength would be pointing vertically downwards so the xv.5 would be subtracted from 127.4 and total forcefulness would be 111.9 Due north?

OlderDan

right, the tension force would be pointing vertically downwards so the 15.five would be subtracted from 127.iv and full forcefulness would exist 111.9 North?

The vertical component of the tension interim where the rope is attaced to the mat is upwardly, in the same direction as the normal. The normal force is upward, opposing the weight.

T_y + Northward = mg

Then yes, the normal force is the weight minus the vertical tension component.

mawalker

gotcha...i'k even so kinda lost on where to go from this point at present

OlderDan

gotcha...i'thou withal kinda lost on where to go from this point now

From the normal strength you can discover the friction, which opposes the horizontal component of the tension in the rope. The resultant of those two is a net horizontal force that accelertes Paul. You can find the acceleration, just the trouble asks yous to utilise piece of work and energy to notice the speed after a certain altitude. What is the net work washed by the forcefulness in the horizontal direction?

mawalker

ok so to notice the forcefulness of friction i took the coefficient of friction .190 times the normal force to give me 21.26 Due north. The tension in the horizontal management would be xv.five N, which would give me 36.7 N as the internet piece of work in the horizontal management.

OlderDan

ok so to find the force of friction i took the coefficient of friction .190 times the normal force to give me 21.26 Northward. The tension in the horizontal direction would exist 15.five N, which would give me 36.7 N as the net work in the horizontal direction.

That is non the horizontal component of the tension. That is the vertiacl compoonent. When yous do detect the horizontal component, recall that friction opposes the motility.