Oh yes, #17. It's a beautiful problem, don't you think? So simple and yet ...
There is a rectangular block and a funny-shaped bracket with a pulley. The block is not supposed to slide on the bracket, so you will use the coefficient of static friction between them, and there is no friction between the bracket and the "frictionless surface". The bracket is pulled with a force of F, which is the tension in the string. I will call it T.
If you set up force diagrams (free body diagrams) and apply Newton's 2nd law, you should be able to get 2 equations with 2 unknowns (a and T), so you can solve for the acceleration (get T= for both equations, and then put the equations T=T and solve for a). Maximum acceleration will occur just before the block slides, so you will use the static friction equality Ff=μN which is valid just before the block slides.
Look at the forces acting on the block. There is tension (call it T) to the left, and there is friction (opposing the direction of motion) to the right, as well as Fg (Fg=m*g) downward and the normal force N upward. The block's acceleration will be to the right if it moves with the bracket. In this case there are no other forces upward or downward and it is not on an incline so (Newton's 2nd law applied to the vertical forces): N-Fg=ma=0 and so N=Fg. Now apply Newton's 2nd law to the horizontal forces on the block: ∑F=Ff-T=ma. I chose the positive direction to be to the right because the block will accelerate to the right. Solve for T: T=Ff-ma.
Now look at the forces on the bracket. There are 2 parts of the string pulling to the right and (from Newton's 3rd law: equal and opposite forces) the same frictional force from the block pulling to the left. The bracket accelerates to the right. Set up Newton's 2nd law: ∑F=T+T-Ff=ma. Solve for T. Remember that the Ff is the same Ff that you used for the block. Remember that the mass in the ma term is the bracket's mass.
Now you have 2 equations that start T=... Put T=T to get an expression (equation) that has known values and acceleration a. The acceleration of the block is the same as the acceleration of the bracket, so you can solve for a.
Thanks Dr. Winters! I was able to get the correct answer by following your directions. I have one question though... In regards to solving for the forces acting on the bracket, you wrote "Remember that the mass in the ma term is the bracket's mass". Why isn't the mass of the block accounted for in this calculation too? (In other words, why wouldn't the masses be added?) I assumed that the question was asking for the acceleration of the whole system, which would include both objects.
When you apply Newton's 2nd law ∑F=ma, the m on the right side is the mass of what you are looking at. So if you are looking at all the forces acting on the bracket, then the mass would be the mass of the bracket. If, on the other hand, you were looking at all the forces acting on the bracket+block system then the mass would be the mass of the bracket+block system. In this problem we were looking at the forces acting on the bracket by itself.
I suppose we could have looked at all the forces acting on the bracket+block system (did anybody out there do that?), but since the bracket+block system would give you one equation that had both the tension T and the acceleration a in it, and you didn't know either, you would eventually need to look at another system: either the bracket by itself or the block by itself.
4 comments:
Oh yes, #17. It's a beautiful problem, don't you think? So simple and yet ...
There is a rectangular block and a funny-shaped bracket with a pulley. The block is not supposed to slide on the bracket, so you will use the coefficient of static friction between them, and there is no friction between the bracket and the "frictionless surface". The bracket is pulled with a force of F, which is the tension in the string. I will call it T.
If you set up force diagrams (free body diagrams) and apply Newton's 2nd law, you should be able to get 2 equations with 2 unknowns (a and T), so you can solve for the acceleration (get T= for both equations, and then put the equations T=T and solve for a). Maximum acceleration will occur just before the block slides, so you will use the static friction equality Ff=μN which is valid just before the block slides.
Look at the forces acting on the block. There is tension (call it T) to the left, and there is friction (opposing the direction of motion) to the right, as well as Fg (Fg=m*g) downward and the normal force N upward. The block's acceleration will be to the right if it moves with the bracket. In this case there are no other forces upward or downward and it is not on an incline so (Newton's 2nd law applied to the vertical forces): N-Fg=ma=0 and so N=Fg. Now apply Newton's 2nd law to the horizontal forces on the block: ∑F=Ff-T=ma. I chose the positive direction to be to the right because the block will accelerate to the right. Solve for T: T=Ff-ma.
Now look at the forces on the bracket. There are 2 parts of the string pulling to the right and (from Newton's 3rd law: equal and opposite forces) the same frictional force from the block pulling to the left. The bracket accelerates to the right. Set up Newton's 2nd law: ∑F=T+T-Ff=ma. Solve for T. Remember that the Ff is the same Ff that you used for the block. Remember that the mass in the ma term is the bracket's mass.
Now you have 2 equations that start T=... Put T=T to get an expression (equation) that has known values and acceleration a. The acceleration of the block is the same as the acceleration of the bracket, so you can solve for a.
"And the rest is just algebra." Good luck.
Thanks Dr. Winters! I was able to get the correct answer by following your directions. I have one question though... In regards to solving for the forces acting on the bracket, you wrote "Remember that the mass in the ma term is the bracket's mass". Why isn't the mass of the block accounted for in this calculation too? (In other words, why wouldn't the masses be added?) I assumed that the question was asking for the acceleration of the whole system, which would include both objects.
When you apply Newton's 2nd law ∑F=ma, the m on the right side is the mass of what you are looking at. So if you are looking at all the forces acting on the bracket, then the mass would be the mass of the bracket. If, on the other hand, you were looking at all the forces acting on the bracket+block system then the mass would be the mass of the bracket+block system. In this problem we were looking at the forces acting on the bracket by itself.
I suppose we could have looked at all the forces acting on the bracket+block system (did anybody out there do that?), but since the bracket+block system would give you one equation that had both the tension T and the acceleration a in it, and you didn't know either, you would eventually need to look at another system: either the bracket by itself or the block by itself.
Okay, I see. Thank you!
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