Calculate Coefficient of Kinetic Friction

[sweetdaisy186]

Hey guys!

I can't seem the get the right answer for the problem below. I'm really close, but I have no idea what I'm doing wrong. I've attached the question and my work. Any suggestions would be greatly appreciated. Thanks!

 

[andrevdh]

The tensions in the string at the two ends are the same so using just T for the magnitude would make life easier, just like you use a for the acceleration of the two masses.

The second mass is accelerating downwards. Choosing down as positive the eom comes to

$$W_2 - T = m_2a$$

your problem arose when you choose the same symbol for the acceleration in the two cases, but the acceleration direction differs as you set your second equation up, requiring a $-a$ for the second case since the problem gives us this information.

 

[sweetdaisy186]

I'm confused. I thought the tension would be the same for both of them. I used T-W2 = M2 * a and then solved for T. Then I set the tensions for both boxes equal to each other. I don't understand what you are saying about the acceleration though. Is that supposed to be part of my free body diaghram?

 

[andrevdh]

When setting up the force equation according to Newton's second law we normally use the symbols for the magnitude and include a sign for the direction. When the same symbol is used in two different equations problems might arise if one do not keep the directions in mind. Your equation for the second mass assumes that it will accelerate upwards since you chose upwards as positive. We know that W2 must be bigger than T to accelerate m2 downwards. So naturally chosing downwards as positive will result in m2a giving a positive value for a. Your choice gives a negative value for a in this case. But for m1 your a will be positive. So the symbol a takes on two different meanings in the two equations. Which will give the incorrect mathematical conclusion. The moral of the story? Choose your positive direction in the direction of the acceleration of the system, or at least so that the connected components will accelerate in the same direction! Not the one part left (m1) and the other part up (m2). Not even the maths can then save you from a disaster.

 

[andrevdh]

I'm confused. I thought the tension would be the same for both of them. I used T-W2 = M2 * a and then solved for T. Then I set the tensions for both boxes equal to each other. I don't understand what you are saying about the acceleration though. Is that supposed to be part of my free body diaghram?

I was just worried for a moment about the tension since you have a T1 and T2 in your equations, but in your derivation you seem to know that they are the same.

As far as the acceleration is concerned you can ignore all that I said if you are willing to change your formula for m2 to

$$T - W_2 = - m_2 a$$

since m2 is accelerating downwards and you chose upwards as positive.

 

[Doc Al]

acceleration constraint

I don't understand what you are saying about the acceleration though. Is that supposed to be part of my free body diagram?

What andrevdh has explained nicely is sometimes called the "acceleration constraint" for the system. In essence, you need to fully take into consideration the fact that the two masses are connected by a string. You already realized that the magnitude of their accelerations are equal, but you also need to reflect the relationship between the direction of their accelerations in your equations. If mass 1 moves down, mass 2 must move left, because they are attached! As andrevdh explained, if you choose your sign convention properly, your equations will work out much easier for you.

You must identify the "acceleration constraint" anytime you have things attached by strings and pulleys.