A series of blocks is connected to a pulley in the manner depicted below. Find the minimum mass that block C must have in order to keep block A from sliding.

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Presentation transcript:

A series of blocks is connected to a pulley in the manner depicted below. Find the minimum mass that block C must have in order to keep block A from sliding on the table top. ss A C B m A =4 Kg m B =2 Kg  s =0.2 (coefficient of static friction) The pulley and rope are to be considered massless. m A =4 Kg m B =2 Kg  s =0.2 (coefficient of static friction) The pulley and rope are to be considered massless. y x

Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Reflection

1. Which physics principle should we use to solve this problem. AA) Work-Energy Theorem BB) Newton’s 2nd Law CC) Law of conservation of energy

Choice A Incorrect Since the blocks are not in motion, there is no work being done. The work-energy theorem will not help us solve this problem.

Since we are considering a situation where the blocks are not slipping on the table and the hanging block is not moving up or down, the system is in static equilibrium. Newton’s Laws tell us that the sum of the forces on an object or the entire system is zero for mechanical equilibrium. In component form in two dimensions, the conditions are: Choice B Correct

The law of conservation of energy is of no use here. Try again. Remember that none of the blocks are in motion in our situation. Choice C Incorrect

2. Let’s begin by analyzing the hanging block B. Which of the following free body diagrams correctly depicts all of the forces acting on block B? A)B)C)A)B)C) mBgmBg T mBgmBg m A g+m C g T mBgmBg T= tension g=9.8m/s 2 T= tension g=9.8m/s 2

There are no forces acting on the hanging block in the positive or negative x-direction. This diagram correctly shows the only two forces acting on the block. Choice A Correct

The weight of blocks A and C do not act on the hanging block directly. They pull on the rope that is connected to block B. The only forces that act on block B are the tension and the block’s weight. Choice B Incorrect

Choice C Incorrect There are no forces acting on the hanging block in the positive or negative x-direction.

3. Using the free body diagram for block B, and applying Newton’s 2nd Law we get which one of the following relations? A)B)C)A)B)C)

Choice A Correct Using Newton’s 2nd Law and our free body diagram we find: After writing down Newton’s 2nd Law resolved into x and y components, we only find a useful equation from the y-component, because none of the forces have x- components.

Choice B Incorrect Using Newton’s 2nd Law and our free body diagram we find: The two forces are subtracted because they act in opposite directions.

Check the free body diagram that we found to be correct in the previous question and try again. Choice C Incorrect

4. Let’s now analyze blocks A and C. Which of the following free body diagrams correctly depicts all of the forces acting on blocks A and C if we treat them as one system? A)B)C)A)B)C) TFfFf (m A -m C )g N N N FfFf T (m A +m C )g FfFf T F f =frictional force N=normal force F f =frictional force N=normal force

The directions of the forces of friction and tension should be switched. The force of static friction opposes the sliding which would be to the right if block C were not massive enough. The rope is attached to block A on the right side, so the tension would act away from the block in the positive x-direction. Choice A Incorrect

This diagram correctly depicts all of the forces acting on the system of blocks A and C. Choice B Correct

The combined weight W of blocks A and C should be considered. We are treating them together as one system. Choice C Incorrect W=(m A -m C )g W=(m A +m C )g

5. Use the free body diagram from the previous question. Which of the following relations do we get by applying Newton’s 2nd Law to blocks A & C together as one system? (Resolve the forces into x and y components.) A)B)C)A)B)C)

Choice A Correct Using Newton’s 2nd Law and our free body diagram we find:

Choice B Incorrect Using Newton’s 2nd Law and our free body diagram we find:

The weight of the system of blocks A and C has no x-component. Choice C Incorrect

6. Use the information from the last question, to find another way to express T in terms of known quantities and the quantity that we are interested in finding, which is the minimum mass (m C ) that block C can have to keep the system in equilibrium. Which one of the following expressions is correct? A)B)C)A)B)C) Note: The force of static friction that will keep the blocks from slipping must satisfy the following condition: We want to find the minimum mass of block C that allows this condition to be satisfied. This occurs when: Note: The force of static friction that will keep the blocks from slipping must satisfy the following condition: We want to find the minimum mass of block C that allows this condition to be satisfied. This occurs when:

Choice A Incorrect Combine several equations that we have found in previous questions,

Choice B Incorrect Combine several equations that we have found in previous questions.

Choice C Correct We combine several equations that we have found in previous questions.

7. We now have two equations for the tension in the rope T. If we equate the two, we can solve for the minimum mass of block C. What is the minimum mass that block C must have in order to keep block A from sliding on the table top? AA) m C =10kg BB) m C =6kg CC) m C =4kg

Choice A Incorrect Please try again. You should find the following expression for the minimum mass of block C: Begin by equating these two expressions for tension.

Choice B Correct The force of friction on the combined mass (m A +m C ) is balanced by the tension force. Block C must have a mass of at least 6 kg in order to keep the other blocks from moving.

Choice C Incorrect Please try again. You should find the following expression for the minimum mass of block C: Begin by equating these two expressions for tension.

Reflection Questions If the coefficient of static friction was lower than 0.2, what would happen to blocks A and C? If the coefficient of static friction was higher than 0.2, what would happen to blocks A and C? More

Reflection Question: How do we know that the relation: gives the minimum mass required to keep the blocks from slipping? Help

Using the inequality we find: This means that the mass of block C must be greater than or equal to 6 kg in order to keep the blocks from slipping. Obviously, 6kg is the minimum mass to satisfy this condition, so if we just use the equal sign originally, we find the minimum mass. This means that the mass of block C must be greater than or equal to 6 kg in order to keep the blocks from slipping. Obviously, 6kg is the minimum mass to satisfy this condition, so if we just use the equal sign originally, we find the minimum mass.