The problem involves a 15 kg block on a 22-degree incline, held stationary by a string, with a coefficient of friction of 0.12. The goal is to determine the minimum tension in the string while considering the forces acting on the block.
The discussion has explored various interpretations of the forces involved, particularly the gravitational forces and friction. Some participants have offered guidance on considering the maximum static friction and its impact on tension. There is an ongoing examination of the calculations and assumptions made regarding the forces acting on the block.
Participants note the importance of ensuring calculations are performed in degrees rather than radians, and there is a focus on the equilibrium of forces acting on the block.
Yes that will be necessary, but then what? Consider a free body diagram, what are all the forces acting on the block? Two of them are variable.Ryan Lau said:Not certain, but if I calculate the Fgparallel portion, will I be able to calculate my Tension Force?
Well, yes. But not for your reasoning. The reason is that the force of static friction is variable. μsFn only represents the maximum force that static friction can apply, but it is possible for static friction to apply less or even no force at all.Ryan Lau said:Well, I notice that both Tension Force and Force Friction are along the same path, therefore I believe those are the two variables?
I noticed you mentioned static friction.Nathanael said:Well, yes. But not for your reasoning. The reason is that the force of static friction is variable. μsFn only represents the maximum force that static friction can apply, but it is possible for static friction to apply less or even no force at all.
You're asked to find the minimum tension in the rope. When do you think the force of tension will be minimum?
When the static friction is zero? When the static friction is maximum? Or somewhere in between?
It can be somewhere in between. Or it can be maximum. Or it can be zero. All of these are valid possibilities. But for each of these options, the force of tension will be different.Ryan Lau said:Since the object is at rest, I believe that the static friction is somewhere in between.
Nathanael said:It can be somewhere in between. Or it can be maximum. Or it can be zero. All of these are valid possibilities. But for each of these options, the force of tension will be different.
When will the force of tension be minimum? Are you saying tension is minimum when the force of friction is somewhere in between it's minimum (zero) and it's maximum?
Good. So now find the minimum force of tension. (That is, find the tension needed to keep the object stationary when static friction is applying it's maximum force.)Ryan Lau said:Tension force is minimum when force of friction is at it's maximum.
Nathanael said:Good. So now find the minimum force of tension. (That is, find the tension needed to keep the object stationary when static friction is applying it's maximum force.)
Nathanael said:How do you get Fgravity parallel=64.7 N and Fgravity perp.=147 N?
Nathanael said:Ok you have the right idea, but 64.7 is not the right number.
Can you finish solving for the minimum tension now?
This is not true... You need to make sure your calculator is in "degrees mode" and not radians.Ryan Lau said:147cos68 = 64.70102431 N
Yes this is correct.Ryan Lau said:Should the opposite of it not equal to both the Tension Force and Frictional Force combined? I thought that is how equilibrium works.
Yes this is also correct.Ryan Lau said:My assumptions are that (0.12 Mew * Normal Force) is my maximum Frictional Force?
I'm not sure I understand what you're confused about. Are you having trouble finding the normal force?Ryan Lau said:The issue is also found when I try to find the adjacent for the 22 degree triangle.
Oh my heavens...Nathanael said:This is not true... You need to make sure your calculator is in "degrees mode" and not radians.Yes this is correct.Yes this is also correct.I'm not sure I understand what you're confused about. Are you having trouble finding the normal force?
Good job, it is correct.Ryan Lau said:Oh my heavens...
Mr. Nathanael, I am terribly sorry... This entire time I was using radians!
Brief overview.
Force of Gravity Parallel = 147cos68 = 55.0672 N
Force of Gravity Perpendicular = 147cos22 = 136.2960 N
Force of Friction = (Mew * Normal Force) = (0.12)(136.2960) = 16.35552
MINIMUM TENSION FORCE = Force of Gravity Parallel - Force of Friction = (55.0672) - (16.35552) = 38.71168 = 39 N