Find Force, Net Force, Acceleration & Final Speed

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Homework Help Overview

The problem involves a box being pushed up a ramp with a constant force at an angle, requiring the calculation of various forces and kinematic quantities, including friction, net force, acceleration, and final speed. The subject area pertains to dynamics and friction in physics.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculation of the normal force and frictional force, with some questioning the signs and components of the forces involved. There are attempts to clarify the setup and the relationships between the forces acting on the box.

Discussion Status

There is ongoing dialogue about the correctness of the calculations and assumptions made regarding the forces. Some participants have offered corrections and alternative perspectives on the force components, while others express differing views on the interpretation of the problem statement.

Contextual Notes

Participants note discrepancies in calculations and assumptions, particularly regarding the direction of forces and the setup of the problem. The original poster expresses urgency due to an upcoming test, indicating a time constraint on resolving these issues.

shar_p
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Homework Statement


I have a test on this tomorrow .. please help..
A box, initially at rest, is pushed up a ramp by a constant force acting 20 degrees above the ramp’s surface (pushing into the ramp from behind the box). The magnitude of the force is 50.0 N. The coefficient of friction is 0.30. The box has a mass of 5.0 kg and the ramp has an angle of 25 degrees. (a)Find the force of friction. (b)Find the net force. (c)Find the acceleration. (d)Find the final speed of the box after 6.0 s. Find how far up the ramp the box will travel in 6.0 s.


Homework Equations


2 different angles th1 and th2
Ff = ?, Fnet = ?
Calc Fn first then from Ff = (Mu)Fn, calc Ff, that would be (a)
Sum of Fx would be (b)
Using Sum of Fx = m. a we can get acc
using d = 1/2 a t ^ 2 (because vi = 0) we get d at 6 sec
using vf^2 = vi^2 + 2ad we can get vf
But I am not getting the correct value of Ff

The Attempt at a Solution


Sum of Fx = Fpx - Ff - Fgx = ma
Sum of Fy = Fn - Fpy - Fgy = 0
Fn = 50sin20 + 5 cos20 = 17.10
Ff = (0.3) Fn = 6.54 but the correct ans is 18 N and so everything else is going wrong.. pleaase help.
 
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Fn = 50sin20 + 5*9.8*cos25.
Now find Ff
 
shar_p said:
Sum of Fy = Fn - Fpy - Fgy = 0
You have the wrong sign for Fpy; it acts in the same direction as the normal force.
Fn = 50sin20 + 5 cos20 = 17.10
In addition, as rl.bhat described, your calculation of Fgy is incorrect.

Thus this value for Fn is wrong.
 
Thanks for the reply. rl.bhat thanks for catching my silly mistake. Glad this problem was not on the test. Though I don't agree with Doc Al that the sign -Fpy is wrong in
Sum of Fy = Fn - Fpy - Fgy = 0
because Fpy is downward because the Fp is pushed up the ramp and because of that Fn = Fpy+Fgy and with 50sin20 + 5*9.8*cos 25 I get the correct answer. Thanks
 
shar_p said:
Though I don't agree with Doc Al that the sign -Fpy is wrong in
Sum of Fy = Fn - Fpy - Fgy = 0
because Fpy is downward because the Fp is pushed up the ramp and because of that Fn = Fpy+Fgy and with 50sin20 + 5*9.8*cos 25 I get the correct answer.
I see what you mean now, but your problem statement said:
shar_p said:
A box, initially at rest, is pushed up a ramp by a constant force acting 20 degrees above the ramp’s surface (pushing into the ramp from behind the box).
I guess you meant that the force was pointing in a direction 20 degrees below the ramp surface. My bad. :redface:
 

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