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

AI Thread Summary
A box weighing 5.03 kg is at rest on a 30-degree incline, and the problem involves calculating the constant force needed to push it to the top in 4.45 seconds, considering both static and kinetic friction. The calculated acceleration is 0.4425 m/s², leading to a net force of 2.225 N required to maintain this acceleration. The confusion arises regarding the force needed to overcome static friction before the box begins to move, as only kinetic friction applies once it is in motion. It's emphasized that the force must first exceed static friction to initiate movement, and then ensure it is sufficient to maintain the desired acceleration. The discussion highlights the need to clarify calculations and understand the transition from static to kinetic friction in this context.
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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