Question regarding tension and friction

[tristan_fc]

Okay, well I've been working on this assignment forever now, and I have it all correct, but one last bit. Here's the problem.

Two blocks connected by a rope of negligible mass are being dragged by a horizontal force F. Suppose that F = 66.0 N, m1 = 10.0 kg, m2 = 20.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.105.

The first part asked for the Tension of the rope between the blocks. I found that to be 22N. I have no idea how, because I had to find the acceleration to find the tension. I found the acceleration to be(incorrectly according to webassign) 2.095m/s^2.

So my question is how do I find the acceleration? I've done it every way I could see, and got 2.095 every time. Any ideas?

One more thing, this problem is due midnight central time.

 

[Sonty]

First of all 66>(10+20)*10*0.105 so the system accelerates forward.
Both blocks will move with the same acceleration (obviously) so you just have to balance the forces on each block and write Newton's second. You'll get a system of two equations with two unknowns: a and T. Like this:
m₁a=F-m₁gμ-T

 

[M98Ranger]

Hi there,

It seems like you have already done the majority of the problem correctly. The tension of the rope between the blocks can be found by using Newton's second law, which states that the net force on an object is equal to its mass multiplied by its acceleration (F=ma). In this case, the net force is the horizontal force F, and the mass is the combined mass of the two blocks (10.0 kg + 20.0 kg = 30.0 kg). So, the tension can be calculated as T = (30.0 kg)(2.095 m/s^2) = 22 N.

As for finding the acceleration, you can use the formula F=ma again, but this time the net force is the difference between the applied force F and the force of friction. The force of friction can be calculated as the coefficient of kinetic friction (0.105) multiplied by the normal force (which is equal to the weight of the block, m*g, where g is the acceleration due to gravity, 9.8 m/s^2). So, the equation would be F-Ff=ma, where Ff = (0.105)(10.0 kg)(9.8 m/s^2) + (0.105)(20.0 kg)(9.8 m/s^2) = 31.5 N. Plugging in the given values, we get (66.0 N - 31.5 N)/30.0 kg = a, which gives us an acceleration of 1.45 m/s^2.

I hope this helps and clarifies any confusion you had. Good luck with your assignment and don't forget to submit it before midnight central time!

 

[M98Ranger]

Hi there,

First of all, great job on finding the tension of the rope between the blocks! To find the acceleration, you can use Newton's second law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration (F=ma). In this case, the net force is the horizontal force F and the force of friction, which is equal to the coefficient of kinetic friction multiplied by the normal force (Ff=μN). The normal force is equal to the weight of the block, which is mg. So, you can set up the equation as follows:

F-Ff=ma

Substituting the given values, we get:

66.0 N - (0.105)(10.0 kg)(9.8 m/s^2) - (0.105)(20.0 kg)(9.8 m/s^2) = (10.0 kg + 20.0 kg)a

Solving for a, we get:

a = 0.668 m/s^2

This is the correct acceleration, not 2.095 m/s^2. Make sure to double check your calculations and units to avoid any errors. Also, don't forget to include the negative sign for the force of friction, as it acts in the opposite direction of the applied force.

I hope this helps! Good luck with your assignment and don't forget to submit it before the deadline.