How Do You Calculate the Weight of a Block on an Inclined Plane?

[Ceshe000]

Homework Statement

A block is placed on a inclined plane. The incline of the plane is 21° .A 24 N force is required to push the block up the incline with constant velocity. The force is applied at an angle of 15° as shown in the picture. The coefficient of kinetic friction (μk) is 0.26. What is the weight of the block? Answer in units of N. The acceleration of gravity is 9.8 m/s/s.

Homework Equations

Force Kinetic Friction = μk * Normal Force

The Attempt at a Solution

The farthest I got in solving the problem was to come up with the force of gravity being 186.5076 N, however I am not entirely sure that this is correct.
To get this I listed the components for each force:
Fn= < 0, Fn >
Fg= < -sin(21)Fg, -cos(21)Fg >
Fpush= < cos(15)Fp, -sin(15)Fp >
Fk= < -μkFn, 0 >
Then is added all the x and y parts together:
Fx:
-sin(21)Fg + cos(15)Fp - μkFn = 0
Fy:
Fn - cos(21)Fg - sin(15)Fp = 0
Then I solved for Fn for each:
Fx:
(cos(15)Fp - sin(21)Fg) / μk = Fn
Fy:
sin(15)Fp + cos(21)Fg = 0
Then I plugged in one Fn for the other:
(cos(15)Fp - sin(21)Fg) / μk = sin(15)Fp + cos(21)Fg
Then I solved for Fg and I am not sure where I messed up doing so...

 

[gneill]

Please show the details of what you've tried.

 

[gneill]

To get this I listed the components for each force:
Fn= < 0, Fn >
Fg= < -sin(21)Fg, -cos(21)Fg >
Fpush= < cos(15)Fp, -sin(15)Fp >
Fk= < -μkFn, 0 >
Then is added all the x and y parts together:
Fx:
-sin(21)Fg + cos(15)Fp - μkFn = 0
Fy:
Fn - cos(21)Fg - sin(15)Fp = 0
Then I solved for Fn for each:
Fx:
(cos(15)Fp - sin(21)Fg) / μk = Fn
Fy:
sin(15)Fp + cos(21)Fg = 0 ← 0 should be Fn.
Then I plugged in one Fn for the other:
(cos(15)Fp - sin(21)Fg) / μk = sin(15)Fp + cos(21)Fg
Then I solved for Fg and I am not sure where I messed up doing so...

What you've done so far is good. I spotted one typo (red comment) which didn't affect the later step. I hope that you understand that equating the two expressions for the normal force is really a subtle way of enforcing the given constraint that the acceleration is zero. A less subtle way would be to declare that the net force in the x-direction (along the slope) is zero and write the sum accordingly. You'll arrive at the same relationship that you did in your last step shown.

(cos(15)Fp - sin(21)Fg) / μk = sin(15)Fp + cos(21)Fg

which looks good to me. So it seems that your problem now is solving the above for Fg as you've surmised and is a matter of algebra. Can you show us how you've tried to isolate Fg in the above?

 

[Ceshe000]

What you've done so far is good. I spotted one typo (red comment) which didn't affect the later step. I hope that you understand that equating the two expressions for the normal force is really a subtle way of enforcing the given constraint that the acceleration is zero. A less subtle way would be to declare that the net force in the x-direction (along the slope) is zero and write the sum accordingly. You'll arrive at the same relationship that you did in your last step shown.

(cos(15)Fp - sin(21)Fg) / μk = sin(15)Fp + cos(21)Fg

which looks good to me. So it seems that your problem now is solving the above for Fg as you've surmised and is a matter of algebra. Can you show us how you've tried to isolate Fg in the above?

Thanks but I already figured it out. But again thank you and the site for the help! Should I delete the thread or post the answer?

 

[gneill]

Thanks but I already figured it out. But again thank you and the site for the help! Should I delete the thread or post the answer?

Threads are never deleted unless there's some extraordinary reason. You can post your answer if you wish, but if you're satisfied with the way things are it's not a requirement. You might wish to mark the thread "solved" to indicate that you're done here (find the "MARK SOLVED" button at the top right of the page).

 

[Ceshe000]

I have solved the problem and the answer is 35.87958 N for anyone that wishes to know.

 

[gneill]

I have solved the problem and the answer is 35.87958 N for anyone that wishes to know.

That's good. Note that your given values had only two significant figures. How many significant figures should you present in your answer?