Max Force on 3.0kg Box to Accelerate: Find the Answer

[allielove13]

Homework Statement

Two boxes sit on a table, connectedn by a massless string. The first is 2.0 kg and the second is 3.0 kg. The coefficient of kinetic friction betweent the boxes and the table is 0.80. The max tensile strength is 1000 N. What is the max force that can be exerted on the 3.0 kg box to make the system accelerate?

Homework Equations

Fn=mg
Fg=mg
Fk=ukFn
Fne=ma?

The Attempt at a Solution

Normal Force on 2.0 kg box:
Fn=mg=2.0(9.8)=19.6 N
Normal Force on 3.0 kg box:
Fn=mg=3.0(9.8)=29.4 N
Total Amount of Friction:
Fk=uk(Fn2.0+Fn3.0)=0.8(19.6+29.4)=39.2 N

Now I am stuck..

 

[BigJDubs]

I don't know what to do from here. Do I use Fnet=ma? I am not sure how to interpret it in this situation.

 

[tomcue]

I would first clarify the question and make sure all the necessary information is provided. The question asks for the maximum force that can be exerted on the 3.0 kg box to make the system accelerate, but it does not specify in which direction the acceleration should occur. Is the question asking for the maximum force in any direction, or only in the direction of the applied force?

Assuming that the question is asking for the maximum force in any direction, we can use the equation F=ma to find the maximum acceleration that can be achieved with the given conditions. Since the system is connected by a string and the boxes are on a table, we can assume that the force will be applied horizontally. Therefore, the maximum force that can be exerted on the 3.0 kg box to make the system accelerate is equal to the maximum static friction force, which is given by μsFn.

We already calculated the normal forces on the boxes to be 19.6 N and 29.4 N, and the total friction force to be 39.2 N. We also know that the maximum static friction force is equal to μsFn, where μs is the coefficient of static friction and Fn is the normal force. Therefore, we can set up the following equation:

μsFn = 39.2 N

Solving for Fn, we get:

Fn = 39.2 N / μs = 39.2 N / 0.80 = 49 N

This is the maximum normal force that can be exerted on the 3.0 kg box to make the system accelerate. Now, using the equation F=ma, we can calculate the maximum acceleration that can be achieved:

F = ma
49 N = (3.0 kg) a
a = 16.3 m/s^2

Therefore, the maximum force that can be exerted on the 3.0 kg box to make the system accelerate is 49 N, and the maximum acceleration that can be achieved is 16.3 m/s^2.

It is also worth noting that the maximum tensile strength of the string, which is given as 1000 N, is not relevant to this question as it is only applicable when the string is being pulled or stretched, not when it is simply supporting the weight of the boxes.