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dioxy186
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Unless otherwise stated, all objects are located near the Earth's surface, where g = 9.80 {\rm m/s^2}.
A person has a choice while trying to push a crate across a horizontal pad of concrete: push it at a downward angle of 29∘ , or pull it at an upward angle 29∘.
If the crate has a mass of 47.0kg and the coefficient of kinetic friction between it and the concrete is 0.740, calculate the required force to move it across the concrete at a steady speed for both situations.
I'm stuck on this problem, I don't know if I have my equations mixed up or I'm calculating the wrong thing.
Fk (is Kinetic Friction)
mg is mass * gravity (weight of the object)
muk (coefficient of kinetic friction - given as 0.74).
N is the Normal force (same as Fn as explained below).
Fn= mg*sin(theta)
Fk= muk*N
plug Fn into N since the coefficient of friction is given.
The frictional force when pulling is µW1:
w-sin(theta)*Fp
460.6 - sin(29)Fp (29 degrees is given - plugged into place of theta).
The frictional force when pushing is µW2:
460.6+sin(29)Fp
Ff(force of friction) Pull:
muk (w - sin(theta)
= (0.74) (47*9.8-sin(29) = 340.8 - 0.359Fp
Ff Push:
(0.74)(47*9.8+sin(29) = 340.8+0.359Fp
The horizontal component is Fp·cos(29°)
Fp*cos(29) = 0.875Fp
The net force when pulling is:
0.875Fp - (340.8 - 0.3588) = 0 -> 0.875Fp+0.3588 Fp -340.8 -> = 1.23 Fp - 340.8
The net force when pushing is:
0.875Fp - (340.8+0.3588) -> 0.875Fp - 0.3588 Fp -340.8 -> = 0.5162Fp - 340.8
Since it's moving at constant speed, does that mean the horizontal component Fp equal to the frictional force in each case?
This is basically where I'm lost at. I try setting Fp = Ff but I'm not getting the correct answers.
A person has a choice while trying to push a crate across a horizontal pad of concrete: push it at a downward angle of 29∘ , or pull it at an upward angle 29∘.
If the crate has a mass of 47.0kg and the coefficient of kinetic friction between it and the concrete is 0.740, calculate the required force to move it across the concrete at a steady speed for both situations.
Homework Equations
I'm stuck on this problem, I don't know if I have my equations mixed up or I'm calculating the wrong thing.
Fk (is Kinetic Friction)
mg is mass * gravity (weight of the object)
muk (coefficient of kinetic friction - given as 0.74).
N is the Normal force (same as Fn as explained below).
Fn= mg*sin(theta)
Fk= muk*N
plug Fn into N since the coefficient of friction is given.
The frictional force when pulling is µW1:
w-sin(theta)*Fp
460.6 - sin(29)Fp (29 degrees is given - plugged into place of theta).
The frictional force when pushing is µW2:
460.6+sin(29)Fp
Ff(force of friction) Pull:
muk (w - sin(theta)
= (0.74) (47*9.8-sin(29) = 340.8 - 0.359Fp
Ff Push:
(0.74)(47*9.8+sin(29) = 340.8+0.359Fp
The horizontal component is Fp·cos(29°)
Fp*cos(29) = 0.875Fp
The net force when pulling is:
0.875Fp - (340.8 - 0.3588) = 0 -> 0.875Fp+0.3588 Fp -340.8 -> = 1.23 Fp - 340.8
The net force when pushing is:
0.875Fp - (340.8+0.3588) -> 0.875Fp - 0.3588 Fp -340.8 -> = 0.5162Fp - 340.8
Since it's moving at constant speed, does that mean the horizontal component Fp equal to the frictional force in each case?
This is basically where I'm lost at. I try setting Fp = Ff but I'm not getting the correct answers.
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