Frictional Force and Acceleration

[LCB]

A 3.80-kg box is sliding across the horizontal floor of an elevator. The coefficient of kinetic friction between the box and the floor is 0.500. Determine the kinetic frictional force that acts on the box when the elevator is

(a) accelerating upward with an acceleration whose magnitude is 2.70 m/s2, and

(b) accelerating downward with an acceleration whose magnitude is 2.70 m/s2.

For both, I started out with Ff=uFn
For (a), I did: Fn + 2.7 = 3.8. Solved for Fn, and of course, my answer was wrong.
For (b), I did: 3.8 + 2.7 = Fn. Again, wrong answer.
I did everything the way my teacher taught me to. I'm incredibly frustrated; I don't understand my mistake(s).

 

[Dorothy Weglend]

A
(a) accelerating upward with an acceleration whose magnitude is 2.70 m/s2, and

For both, I started out with Ff=uFn
For (a), I did: Fn + 2.7 = 3.8. Solved for Fn, and of course, my answer was wrong.

I think gravity is working in the elevator, so you need it in your equations. If your elevator is going up, then it will press harder against the box, so the acceleration will be (2.7 m/s^2 + g). And if the elevator is going down, then it is pressing less, so the total acceleration will be less than g.

And then, you must use F=ma to get the normal force, once you have figured out the a.

I hope this helps you out.
Dot

 

[quicknote]

I can provide some insight on the problem you are facing. First, let's review the basics of frictional force and acceleration. Frictional force is a force that opposes the motion of an object and is caused by the interaction between two surfaces. The coefficient of kinetic friction is a measure of the roughness of the surfaces in contact, and it determines the magnitude of the frictional force.

Now, let's apply this knowledge to the given scenario. The box is sliding across the floor, which means it is experiencing a kinetic frictional force in the opposite direction of its motion. This can be represented by the equation Ff = μkFn, where μk is the coefficient of kinetic friction and Fn is the normal force (perpendicular to the surface) acting on the box.

In part (a), the elevator is accelerating upward with an acceleration of 2.70 m/s2. Since the box is in contact with the floor, it is also experiencing this acceleration. This means that there is an additional force acting on the box, which is the force of acceleration (F=ma). In this case, the normal force acting on the box is not equal to its weight (3.8 kg), as the box is being pushed upward by the floor. Therefore, we need to consider the net force acting on the box in the vertical direction.

Using Newton's second law, we can write the equation for the net force in the vertical direction as ΣF = Fg - Ff - F = ma, where Fg is the weight of the box (mg), Ff is the frictional force, and F is the force of acceleration. Substituting the values, we get:

ma = mg - μkFn - F
2.70(3.8) = 3.8(9.8) - 0.500Fn - F
Fn = 23.4 N

Now, to find the frictional force, we can substitute the value of Fn in the equation Ff = μkFn, which gives us:

Ff = (0.500)(23.4) = 11.7 N

Similarly, for part (b), the net force in the vertical direction can be written as ΣF = Fg + Ff - F = ma. Substituting the values, we get:

ma = mg + μkFn - F
2.70(3.

 

[quicknote]

I would like to clarify that the coefficient of kinetic friction is a measure of the resistance to motion between two surfaces in contact and it is not affected by the acceleration of the object. Therefore, the value of 0.500 remains constant for both parts of the problem.

To determine the kinetic frictional force, we need to use the formula Ff = μFn, where μ is the coefficient of kinetic friction and Fn is the normal force acting on the box. The normal force is the force that the floor exerts on the box perpendicular to its surface.

For part (a), the box is accelerating upward with an acceleration of 2.70 m/s^2. This means that the net force acting on the box is the sum of the kinetic frictional force and the force due to acceleration, which is given by Fnet = ma. Therefore, we can write the equation Ff + ma = Fn. Substituting the values, we get Ff + (3.80 kg)(2.70 m/s^2) = Fn. Solving for Fn, we get Fn = 13.06 N. Now, substituting this value for Fn in the formula Ff = μFn, we get Ff = (0.500)(13.06 N) = 6.53 N. Therefore, the kinetic frictional force acting on the box when the elevator is accelerating upward is 6.53 N.

For part (b), the box is accelerating downward with an acceleration of 2.70 m/s^2. This means that the net force acting on the box is the difference between the kinetic frictional force and the force due to acceleration, which is given by Fnet = ma. Therefore, we can write the equation Ff - ma = Fn. Substituting the values, we get Ff - (3.80 kg)(2.70 m/s^2) = Fn. Solving for Fn, we get Fn = -6.53 N. Now, substituting this value for Fn in the formula Ff = μFn, we get Ff = (0.500)(-6.53 N) = -3.27 N. Therefore, the kinetic frictional force acting on the box when the elevator is accelerating downward is -3.27 N.

In conclusion, the kinetic frictional force acting on the box remains constant at 6.53 N regardless of the direction of acceleration