Element1674
- 29
- 0
Two forces questions - Extremely difficult:(
Problem A: A stack of dinner plates on a kitchen counter is accelerating horizontally at 2.7m/s^2. Determine the smallest coefficient of static friction between the dinner plates that will prevent slippage.
Problem B: A 24kg box is tied to a 14kg box with a horizontal rope. The coefficient of friction between the boxes and floor is 0.32. You pull the larger box forward with a force of 180N at an angle 25deg above the horizontal. Find the acceleration of the boxes and the tension of the rope.
Fnet=ma=sum of all forces
Friction=(u)(Fn)
For problem A, the wording makes no sense. If they are accelerating, are they not slipping? Static friction only applies when the object is at rest, and the plates are not. And I assumed this is how I would solve it:
ma=Ff+Fa
ma-Fa=uFn
(Ma-Fa)/mg
Masses cancel:
a-Fa divided by g = u. But I don't know the applied force. This question is just weird.
For problem B, I did this:
The following is for the vertical direction:
Fnet=0=ma=Fn+Fg+Fa
But I get confused as to which masses to use :/. I originally did this:
Fnet=0=mg=Fn-mg+180sin25 (up taken as positive direction)
Fn=mg-180sin25 WHERE M IS EQUAL TO THE SUM OF BOTH MASSES (they can be treated as a single object)
X direction:
Fnet=ma= some non-zero value = Fa-Ff
Ma=180cos25-u(mg-180sin25)
Divide both sides by mass and solve
Therefore a=1.8m/s^2
And this is awkward because I just got the correct answer... Idk perhaps I experienced a calculator error before. But calculating the tension in the rope, I don't even know where to begin. Isn't it just equal to the net force of the smaller mass?
Homework Statement
Problem A: A stack of dinner plates on a kitchen counter is accelerating horizontally at 2.7m/s^2. Determine the smallest coefficient of static friction between the dinner plates that will prevent slippage.
Problem B: A 24kg box is tied to a 14kg box with a horizontal rope. The coefficient of friction between the boxes and floor is 0.32. You pull the larger box forward with a force of 180N at an angle 25deg above the horizontal. Find the acceleration of the boxes and the tension of the rope.
Homework Equations
Fnet=ma=sum of all forces
Friction=(u)(Fn)
The Attempt at a Solution
For problem A, the wording makes no sense. If they are accelerating, are they not slipping? Static friction only applies when the object is at rest, and the plates are not. And I assumed this is how I would solve it:
ma=Ff+Fa
ma-Fa=uFn
(Ma-Fa)/mg
Masses cancel:
a-Fa divided by g = u. But I don't know the applied force. This question is just weird.
For problem B, I did this:
The following is for the vertical direction:
Fnet=0=ma=Fn+Fg+Fa
But I get confused as to which masses to use :/. I originally did this:
Fnet=0=mg=Fn-mg+180sin25 (up taken as positive direction)
Fn=mg-180sin25 WHERE M IS EQUAL TO THE SUM OF BOTH MASSES (they can be treated as a single object)
X direction:
Fnet=ma= some non-zero value = Fa-Ff
Ma=180cos25-u(mg-180sin25)
Divide both sides by mass and solve
Therefore a=1.8m/s^2
And this is awkward because I just got the correct answer... Idk perhaps I experienced a calculator error before. But calculating the tension in the rope, I don't even know where to begin. Isn't it just equal to the net force of the smaller mass?
