Tension with horizontal kinetic friction

[AnkhUNC]

[SOLVED] Tension with horizontal kinetic friction

Homework Statement

In Figure, block 1 of mass m1 = 1.8 kg and block 2 of mass m2 = 1.3 kg are connected by a string of negligible mass. Block 2 is pushed by force of magnitude 20 N and angle θ = 39°. The coefficient of kinetic friction between each block and the horizontal surface is 0.22. What is the tension in the string?

http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c06/fig06_61new.gif

Homework Equations

The Attempt at a Solution

This is what I end up with but its wrong:
20*cos(39)-20*sin(39)*0.22-
1.3*g*0.22-T=1.3*a

and
T-1.8*g*0.22=1.8*a

Solving for T =

a=T/1.8-g*0.22

20*cos(39)-20*sin(39)*0.22-
1.3*g*0.22-T=1.3*(T/1.8-g*0.22)


(20*cos(39)-20*sin(39)*0.22-
1.3*g*0.22+1.3*g*0.22)*1.3/(1/1.8+1)=T

10.7 N

 

[anathema]

= T



I have read your post and I believe I can provide some assistance. First, I would like to clarify the problem statement. Are the blocks on a horizontal surface or are they on an inclined plane? Also, is the string connecting the blocks horizontal or at an angle? This information will be necessary in order to accurately solve the problem.

Assuming that the blocks are on a horizontal surface and the string is also horizontal, the first step would be to draw a free body diagram for each block. For block 1, the forces acting on it would be its weight (mg) and the tension in the string (T). For block 2, the forces would be its weight (mg), the applied force (20 N) at an angle of 39°, and the tension in the string (T).

Next, we can write out the equations of motion for each block. For block 1, the equation would be ΣF=ma, where ΣF is the sum of all the forces acting on the block. This would give us:

T-m1g=0

For block 2, the equation would be ΣF=ma, where ΣF is the sum of all the forces acting on the block. This would give us:

T-m2g-20cos(39)=m2a

We can also use the fact that the coefficient of kinetic friction is given as 0.22 to write out an equation for the frictional force acting on each block. This would be:

μkN=m1g and μkN=m2g

Where N is the normal force acting on each block.

Now, we can substitute the value of N into the equations for each block to get:

T-0.22m1g=0 and T-0.22m2g-20cos(39)=m2a

We can then solve these equations simultaneously to get the value of T, which would be the tension in the string. I hope this helps. Let me know if you have any further questions or need clarification on any of the steps. Good luck with your problem solving!

 

[Moetasim]

= T

The tension in the string is 10.7 N.
Based on the given information, it seems like you have correctly set up the equations for the tension in the string. However, there may be a mistake in your calculations as the final answer of 10.7 N does not seem to match the given values. I would recommend double-checking your calculations and equations to ensure accuracy. Additionally, it may be helpful to label your equations and variables clearly to avoid any confusion. Overall, your approach to the problem seems correct, but it is important to carefully check your work to arrive at the correct answer.