Mechanics- General motion in a straight line.

[Shah 72]

It's the second question.
Limiting equilibrium by a force of 5N means even friction is acting in the same direction. I don't understand how to calculate. Pls help

 

[topsquark]

After all the comments to put your images right side up you still aren't going to do it?

So. Please fix the diagram. In addition, can you give us the Free Body Diagram? That's always the first place to start.

-Dan

 

[Shah 72]

After all the comments to put your images right side up you still aren't going to do it?

So. Please fix the diagram. In addition, can you give us the Free Body Diagram? That's always the first place to start.

-Dan

Apologies. Iam getting confused on how to calculate acceleration without calculating the coefficient of friction.

 

[Shah 72]

After all the comments to put your images right side up you still aren't going to do it?

So. Please fix the diagram. In addition, can you give us the Free Body Diagram? That's always the first place to start.

-Dan

I got R=25.9 N, I got coefficient of friction =0.3336.
Iam confused how to calculate the direction of the particle first without finding coefficient of friction.
I know the direction is down the slope.

 

[Shah 72]

After all the comments to put your images right side up you still aren't going to do it?

So. Please fix the diagram. In addition, can you give us the Free Body Diagram? That's always the first place to start.

-Dan

I got it. Since the force 30 sin 27 acting downward is greater than the force 5 cos 10 acting upwards, the particle slips down

 

[topsquark]

I got it. Since the force 30 sin 27 acting downward is greater than the force 5 cos 10 acting upwards, the particle slips down

Please sketch a FBD. You didn't include friction, where did the of 30 come from, what about the normal force, etc?

And the problem states that the object is not moving.

-Dan

 

[Shah 72]

Please sketch a FBD. You didn't include friction, where did the of 30 come from, what about the normal force, etc?

And the problem states that the object is not moving.

-Dan
[/QUO

Please sketch a FBD. You didn't include friction, where did the of 30 come from, what about the normal force, etc?

And the problem states that the object is not moving.

-Dan

30N is the weight of the particle. Normal contact force is R+ 5sin10= 30 cos 27, R=25.9
Without finding coefficient of friction, ( since that's the first question)
I thought I would say upward force is smaller than downward force, hence the particle slips down

 

[Shah 72]

Please sketch a FBD. You didn't include friction, where did the of 30 come from, what about the normal force, etc?

And the problem states that the object is not moving.

-Dan

 

[DaalChawal]

Your FBD is wrong

 

[topsquark]

I agree. You didn't put in the static friction force. What direction should it point in? Also, your diagram seems to imply that the weight is directly opposite your normal force. It should be straight down.

Oh, and just to verify: You are using g = 10 m/s^2?

-Dan

 

[Shah 72]

Your FBD is wrong

Can you pls show me the correct FBD diagram.

 

[Shah 72]

I agree. You didn't put in the static friction force. What direction should it point in? Also, your diagram seems to imply that the weight is directly opposite your normal force. It should be straight down.

Oh, and just to verify: You are using g = 10 m/s^2?

-Dan

Since its lifting equilibrium, the friction also should act in the same direction as the upward force. So it acts upwards. I did not put any value for acceleration as I wasn't sure.

 

[Shah 72]

I agree. You didn't put in the static friction force. What direction should it point in? Also, your diagram seems to imply that the weight is directly opposite your normal force. It should be straight down.

Oh, and just to verify: You are using g = 10 m/s^2?

-Dan

Sorry limiting equilibrium

 

[topsquark]

Since its lifting equilibrium, the friction also should act in the same direction as the upward force. So it acts upwards. I did not put any value for acceleration as I wasn't sure.

Friction will act either up the slope or down the slope. It can't act upward.

As for the FBD, so long as you have the weight going straight down you need to decide which direction the friction force will be in. So if friction wasn't there what direction would the object move? Then you set the friction force to be in the opposite direction.

Give it a try and post us.

-Dan

 

[Shah 72]

Friction will act either up the slope or down the slope. It can't act upward.

As for the FBD, so long as you have the weight going straight down you need to decide which direction the friction force will be in. So if friction wasn't there what direction would the object move? Then you set the friction force to be in the opposite direction.

Give it a try and post us.

-Dan

If there is no friction acting then the body would move down as downward force of 30sin 27 is greater than 5 cos 10. So friction according to this question will be acting in the upward direction as its in limiting equilibrium. Is this the correct??
For these type of problems my tutors and textbook haven't taught me how to draw free body diagram. If you can pls show me once with a diagram it will be very very helpful as my concept will be clear and I can apply it to other similar type questions.

 

[topsquark]

Sounds good to me.

Free Body Diagram: You have to have some kind of source you are using, don't you? How can you be assigned a problem if you have no reference to learn how to do the problems with? (I've been wondering that also with the Kinematics problems you've been posting.) To draw an FBD you label the object as a point, then draw all forces that act on that object coming from that point. More or less what you sketched, just without the friction (and calling the normal force "R"? I usually use N.)

-Dan

 

[Shah 72]

Sounds good to me.

Free Body Diagram: You have to have some kind of source you are using, don't you? How can you be assigned a problem if you have no reference to learn how to do the problems with? (I've been wondering that also with the Kinematics problems you've been posting.) To draw an FBD you label the object as a point, then draw all forces that act on that object coming from that point. More or less what you sketched, just without the friction (and calling the normal force "R"? I usually use N.)

-Dan

The free body diagram I have drawn is the way I have been taught. Even in my textbook it shows similar diagrams that I have drawn in its examples. ( may be its drawn slightly at an angle but I draw it with the particle slightly straight so that I can draw a straight vertical line).
Thank you so much for clearing that my diagram isn't wrong.

 

[topsquark]

Thank you so much for clearing that my diagram isn't wrong.

The diagram you posted above is wrong. If you fixed yours then you didn't post it.

And you really should make your weight pointing downward. FBD's, like all diagrams, should be taken literally due to scale issues (since we are only sketching it, after all) but you should at least get things pointing in more or less the correct direction.

-Dan

 

[Shah 72]

The diagram you posted above is wrong. If you fixed yours then you didn't post it.

And you really should make your weight pointing downward. FBD's, like all diagrams, should be taken literally due to scale issues (since we are only sketching it, after all) but you should at least get things pointing in more or less the correct direction.

-Dan

Basically only the friction had to be shown in the diagram which would be upwards correct??

 

[topsquark]

Basically only the friction had to be shown in the diagram which would be upwards correct??

Okay, I see what happened.

We need to work on your descriptions a bit. There is a big difference between "upward" and "up the plane." The friction force should be up the plane.

Other than that you are ready to go. Define a coordinate plane with +x down the slope and +y in the direction of the normal force. Then use Newton's 2nd Law in each coordinate direction.

-Dan

 

[Shah 72]

Okay, I see what happened.

We need to work on your descriptions a bit. There is a big difference between "upward" and "up the plane." The friction force should be up the plane.

Other than that you are ready to go. Define a coordinate plane with +x down the slope and +y in the direction of the normal force. Then use Newton's 2nd Law in each coordinate direction.

-Dan

Sorry about the description. Yeah I meant friction acts up the slope and not vertically upwards.
Thank you so so so much!