Find force to push block up incline with static and kinetic friction in certain time

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Homework Help Overview

The problem involves determining the constant force required to push a box up an inclined ramp, considering both static and kinetic friction. The box is initially at rest, and the scenario includes specific dimensions of the ramp and coefficients of friction.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of acceleration and the implications of static versus kinetic friction on the force required. There is confusion regarding the role of static friction and the need for a constant force during the motion.

Discussion Status

Some participants have offered guidance on the calculations, particularly regarding the distinction between static and kinetic friction. There is ongoing exploration of the implications of the calculated acceleration and whether it aligns with the requirements of the problem.

Contextual Notes

Participants note that the force to overcome static friction must be considered before the box begins to move, and there is uncertainty about the correct interpretation of the average velocity and acceleration calculations.

hairE
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Homework Statement


A 5.03 box sits at rest at the bottom of a ramp that is 8.76 long and that is inclined at 30.0 above the horizontal. The coefficient of kinetic friction is 0.40, and the coefficient of static friction is 0.50.

What constant force , applied parallel to the surface of the ramp, is required to push the box to the top of the ramp in a time of 4.45 ?

Homework Equations


max mu_static * force Normal = Force to budge Block
mu_kinetic * force Normal = Force Friction



The Attempt at a Solution



The force to overcome the static friction is what is confusing me.
To start: Find acceleration.
8.76m / 4.45 s = 1.969 m/s = veloc.
1.969 m/s /4.45 s = .4425 m/s^2 = accel.

F= ma = 5.03*.4425 = 2.225 N = Leftover, or net force to keep the block accelerating up the ramp

(Parallel to plane) ƩF= Push - Force static - force kinetic - Weight = 2.25N
2.25N = Push - mgsin(30) - (mu_k)mgcos(30) - (mu_s)mgcos(30)

BUT*** Force static is only acting until the object budges. And the question is asking for a CONSTANT force applied. I'm confused. PLEASE help! I just need the sum of forces explained to me.
 
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hairE said:
(Parallel to plane) ƩF= Push - Force static - force kinetic - Weight = 2.25N
2.25N = Push - mgsin(30) - (mu_k)mgcos(30) - (mu_s)mgcos(30)
Either static or kinetic friction acts--not both (at the same time).

Since the object is moving, use kinetic friction. (Will the calculated force be enough to overcome static friction?)
 


Hi HairE
hairE said:
8.76m / 4.45 s = 1.969 m/s = veloc.

Are you clear as to what velocity you just calculated? Is it the velocity at some particular time during the motion or is it the overall average velocity? With that in mind, are you sure you calculated the acceleration correctly?
 


Oops... TSny is right.
hairE said:
To start: Find acceleration.
8.76m / 4.45 s = 1.969 m/s = veloc.
1.969 m/s /4.45 s = .4425 m/s^2 = accel.
You'll need to redo this calculation. (In addition to what I already pointed out.)
 


Humm. If the coefficient of static friction is higher than the kinetic friction presumably you have to apply a force at least strong enough to get it moving. Then check that force gets it up the ramp quick enough. If not then even more force is required.
 

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