Find the acceleration reached by each of the two objects shown in figure below if the coefficient of kinetic friction between the 7.00 kg object and the plane is 0.350.
Figure: http://www.webassign.net/sf/p4_49.gif
The answer is: 3.01 m/s^2
How do I reach this answer?
I separated my gravitational force into x and y components.
I know F= mass x acceleration
I found the 7 kg box components to be:
Fy= n-mgcos37=0 and Fx=T-mgsin37-Fk=ma
n=mgcos37 I substituted n as mgcos37 in Fk=n*µk
Fx=T-mgsin37-(mgcos37*µk)=ma
T=mgsin37+(mgcos37*µk)+ma
and the 12 kg box to be:
Fy= T-mg=ma
I substituted T from the equation from the 7 kg box
[m1gsin37+(m1gcos37*µk)+m1a]-m2g=m2a
If i plug in what i know i end up with:
-57.14=5a
and i get a= -11.43 m/s^2
what am i doing wrong ?
Figure: http://www.webassign.net/sf/p4_49.gif
The answer is: 3.01 m/s^2
How do I reach this answer?
I separated my gravitational force into x and y components.
I know F= mass x acceleration
I found the 7 kg box components to be:
Fy= n-mgcos37=0 and Fx=T-mgsin37-Fk=ma
n=mgcos37 I substituted n as mgcos37 in Fk=n*µk
Fx=T-mgsin37-(mgcos37*µk)=ma
T=mgsin37+(mgcos37*µk)+ma
and the 12 kg box to be:
Fy= T-mg=ma
I substituted T from the equation from the 7 kg box
[m1gsin37+(m1gcos37*µk)+m1a]-m2g=m2a
If i plug in what i know i end up with:
-57.14=5a
and i get a= -11.43 m/s^2
what am i doing wrong ?
-
Take each box separately. For the first one we know the downward force is 7 x 9.8 which is less than the second ones downward force(12x 9.8). So the 2nd one is accelerating down and the 1st box, up.
Now we are only concerned about the 1st box.
It has 4 forces acting on it. The tension(applied force). Force of friction. The normal force(perpendicular to the surface). Gravitational force (7 x9.8). Now the easiest way to understand this is to know that all the horizontal (xaxis) forces will add up to zero. And same for the vertical(yaxis) forces.
Now heres the tricky bit. If you draw a free body diagram of the first box you will see that the normal is a vertical force. The tension and the frictional force are horizontal forces acting in opposite directions. But the gravitational force is neither. So we break it down to its components (to get a horizontal and vertical force). The horizontal one(which I will call Fx) will be Fsin theta which is 7*9.8*sin37 and the vertical(Fy) will be 7*9.8*cos37. Now we have identified all the vertical and horizontal forces(and remember that each of them will add up to zero). So the normal force will be equal to Fy which is 54.78. Now we can find the force of friction(Ff) which is coefficeint*normal force, 54.78*0.35=19.17N.
Since the first box is moving up, using newton's 2nd law(and common sense) the tension should be greater than the friction force and Fx. So T-(Fx plus Ff) =ma. which is T -(41.28+19.17)=7a.
This is the final equation for the first box: T-60.45=7a.
Now the second box's equation is easier. the only forces acting on it is the upward tension and downward gravitationl force. Since its accelerating down the gravitational frce is greater than the tension. SO 12*9.8 - T=12a. So the final equation for the second box is 117.6-T=12a.
Now we substitute T from the two equations and we get T=7a + 60.45 for the 1st equation and
117.6-12a=T for the second one. Then we equate them to get 7a+60.45=117.6-12a and solve for a. I got 3.0078m/s^2.
Now we are only concerned about the 1st box.
It has 4 forces acting on it. The tension(applied force). Force of friction. The normal force(perpendicular to the surface). Gravitational force (7 x9.8). Now the easiest way to understand this is to know that all the horizontal (xaxis) forces will add up to zero. And same for the vertical(yaxis) forces.
Now heres the tricky bit. If you draw a free body diagram of the first box you will see that the normal is a vertical force. The tension and the frictional force are horizontal forces acting in opposite directions. But the gravitational force is neither. So we break it down to its components (to get a horizontal and vertical force). The horizontal one(which I will call Fx) will be Fsin theta which is 7*9.8*sin37 and the vertical(Fy) will be 7*9.8*cos37. Now we have identified all the vertical and horizontal forces(and remember that each of them will add up to zero). So the normal force will be equal to Fy which is 54.78. Now we can find the force of friction(Ff) which is coefficeint*normal force, 54.78*0.35=19.17N.
Since the first box is moving up, using newton's 2nd law(and common sense) the tension should be greater than the friction force and Fx. So T-(Fx plus Ff) =ma. which is T -(41.28+19.17)=7a.
This is the final equation for the first box: T-60.45=7a.
Now the second box's equation is easier. the only forces acting on it is the upward tension and downward gravitationl force. Since its accelerating down the gravitational frce is greater than the tension. SO 12*9.8 - T=12a. So the final equation for the second box is 117.6-T=12a.
Now we substitute T from the two equations and we get T=7a + 60.45 for the 1st equation and
117.6-12a=T for the second one. Then we equate them to get 7a+60.45=117.6-12a and solve for a. I got 3.0078m/s^2.