Michael J. answered • 03/07/15
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Effective High School STEM Tutor & CUNY Math Peer Leader
We will use the formula
Ff = μN
where:
Ff = friction force
μk = coefficient of kinetic friction
μs = coefficient of static friction
N = Normal force (perpendicular to surface of contact) = mg
Question 1
We will find the friction force of the crate. Then we will find the net force between the friction force and applied force. Divide that net force by the total mass in the system.
Ff = μkN
Ff = (0.4)(50 kg)(9.81 m/s2)
Ff = 16.2 kg m/s2
Net force = Fapplied - Ff
= 200 N - 16.2 N
= 183.80 N
a = Fnet / (mcrate + mbox)
a = (183.80 kg m/s2) / (50 kg + 10 kg)
a = (183.8kg m/s2) / 60 kg
a = 3.06 m/s2
Question 3
The box does not remain in its initial position because the force of the crate pushes the box when they are in contact.
Question 4
We will use the coefficient of static friction because we want to keep the box from moving.
Fs = μsN
Fs = (0.5)(10 kg)(9.81 m/s2)
Fs = 49.05 N
Brad J.
Also, there is no friction between the floor and the system. So wouldn't the net force on the system in the X-Direction be 200n???
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03/14/15
Michael J.
I should have tried using system of equations that applies the static equilibrium concept. Equation 1 could be the
∑Fx = 0 and equation 2 could be the ∑Fy = 0.
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03/14/15
Brad J.
Yay! I did get the same answer as Micheal for the last part, 49N is the lowest value of F that would keep the smaller box from falling.
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03/15/15
Brad J.
But, that is not what the original question asked for. So, if the big box is exerting a force of 49 on the small box to keep it from falling, you need to work backwards to calculate that "F" value. Find the A of the system via F=MA: 49N=10kg x A --> 4.9m/s^2.
4.9m/s^2=Fnet/Mass total --> 4.9=F/60 --> F=294N
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03/15/15
Brad J.
03/14/15