The analysis steps. Problem Analysis Sub-problem 3 Sub-problem 2 Sub-problem nSub-problem 1.
Problem
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Problem # 1 A block of mass m is pulled, via pulley, at constant velocity along a surface inclined at angle θ . The coefficient of kinetic friction is μ k , between block and surface. Determine the pulling force F . Answer: mg cos θμ k + mg sin θ Problem # 2 Two blocks of mass m and M are hanging off a single pulley, as shown. Determine the acceleration of the blocks. Ignore the mass of the pulley. Hint and answer
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MATH
Transcript of Problem
Problem # 1
A block of massmis pulled, via pulley, at constant velocity along a surface inclined at angle. The coefficient of kinetic friction isk, between block and surface. Determine the pulling forceF.Answer:mgcosk+mgsin
Problem # 2
Two blocks of massmandMare hanging off a single pulley, as shown. Determine the acceleration of the blocks. Ignore the mass of the pulley.
Hint and answer
Problem # 3
Two blocks of massmandMare connected via pulley with a configuration as shown. The coefficient of static friction iss, between block and surface. What is the maximum massmso that no sliding occurs?Answer:maximumm=Ms
Problem # 4
Two blocks of massmandMare connected via pulley with a configuration as shown. The coefficient of static friction iss, between block and surface. What is the minimum and maximum massMso that no sliding occurs?
Hint and answer
Problem # 5
Two blocks of massmandMare connected via pulley with a configuration as shown. The coefficient of static friction iss, between blocks and surface. What is the maximum massmso that no sliding occurs?Answer:Maximumm=Ms/(sincoss)
Problem # 6
Two blocks of massmandMare connected via pulley with a configuration as shown. The coefficient of static friction between the left block and the surface iss1, and the coefficient of static friction between the right block and the surface iss2. Formulate a mathematical inequality for the condition that no sliding occurs. There may be more than one inequality.
Hint and answer
Problem # 7
A block of massmis pulled, via two pulleys as shown, at constant velocity along a surface inclined at angle. The coefficient of kinetic friction isk, between block and surface. Determine the pulling forceF. Ignore the mass of the pulleys.
Hint and answer
Problem # 8
A block of massmis lifted at constant velocity, via two pulleys as shown. Determine the pulling forceF. Ignore the mass of the pulleys.
Hint and answer
Problem # 9
A block of massMis lifted at constant velocity, via an arrangement of pulleys as shown. Determine the pulling forceF. Ignore the mass of the pulleys.
Hint and answer
The hints and answers for these pulley problems will be given next.
Hints And Answers For Pulley Problems
Hint and answer for Problem # 2
This is called the Atwood machine and is commonly used for demonstration in physics classes.
Apply Newton's second law to the block on the left. We haveMgT=Ma(taking the downward direction as positive). Apply Newton's second law to the block on the right. We havemgT= -ma(the acceleration of the two blocks have opposite signs, since one moves up and the other moves down). Combine these two equations and we can find an expression for the acceleration of the blocks.
Answer:a= (Mm)g/(M+m)
Hint and answer for Problem # 4
For the maximum massM, the block is on the verge of sliding down the incline. This means thatMgsinTMgcoss= 0, whereTis the tension in the rope. SinceT=mg, we can calculate the maximumMfrom the previous equation.
For the minimum massM, the block is on the verge of sliding up the incline. This means thatMgsinT+Mgcoss= 0, whereT=mg. We can calculate the minimumMfrom the previous equation.
Answer:MinimumM=m/(sin+coss), MaximumM=m/(sincoss)
Hint and answer for Problem # 6
This is a challenging problem! It took me a while to figure this one out!
At some angle1>max1blockMwill slide down on its own if there is no rope attached. Similarly, at some angle2>max2blockmwill slide down on its own if there is no rope attached.
It is known thatmax1= atan(s1) andmax2= atan(s2).
If1max1and2max2then no sliding occurs. There are three more cases to consider.
Case 1:
1>max1and2max2.
Apply the equilibrium equation to blockMin which it is on the brink of sliding down. We have:Mgsin1Mgcos1s1Tmin1= 0, whereTmin1corresponds to the minimum rope tension preventing blockMfrom sliding down. (Note that the system naturally "settles" such that the rope tensionTrequired to stop the block from sliding down is the minimum possible amount). ForTmax2.
This is the same as case 1, by symmetry. Hence, the final inequality for this case is:
msin2mcos2s2Msin1+Mcos1s1
Case 3:
1>max1and2>max2.
The blocks will slide together in one direction or the other. To determine the direction we must first calculate the net force pulling down on each block along their respective inclines, as a result of gravity. We do this as follows:
For blockM,Fnet1=Mgsin1Mgcos1s1. AndFnet1> 0 since1>max1.
For blockm,Fnet2=mgsin2mgcos2s2. AndFnet2> 0 since2>max2.
We now have three sub-cases to consider. The final inequalities for this case will be given within these three sub-cases, as follows.
Case 3A:
IfFnet1=Fnet2the blocks will not slide.
Case 3B:
IfFnet1>Fnet2, thenFnet1mgsin2+mgcos2s2for no sliding. Note thatFnet1is equal to the rope tension, and this rope tension is the minimum required to prevent blockMfrom sliding down the incline.
Hence, for no sliding:Msin1Mcos1s1msin2+mcos2s2
Case 3C:
IfFnet2>Fnet1, thenFnet2Mgsin1+Mgcos1s1for no sliding. Note thatFnet2is equal to the rope tension, and this rope tension is the minimum required to prevent blockmfrom sliding down the incline.
Hence, for no sliding:msin2mcos2s2Msin1+Mcos1s1
We are done!
Hint and answer for Problem # 7
Apply the condition of static equilibrium to the block. We have 2Fmgsinmgcosk= 0. The term 2Fcomes from a force analysis in which we see that there are two segments of rope pulling equally on the block. We then solve this equation forF.
Answer:F= (1/2)mg(sin+kcos)
Hint and answer for Problem # 8
Apply the condition of static equilibrium to the block. We have 2Fmg= 0. The term 2Fcomes from a force analysis in which we see that there are two segments of rope pulling equally on the block. We then solve this equation forF.
Answer:F=mg/2
Hint and answer for Problem # 9
Upon close inspection we see that the bottom two pulleys are held up by four segments of rope. The tension in the rope is assumed equal throughout its length (a good assumption for ropes in general since they weigh little). Three of the four rope segments are vertical while the remaining rope segment is at a small angle with the vertical. But for ease of calculation we can treat it as being exactly vertical. Since we are ignoring the mass of the pulleys, the tension in the four rope segments must equal the weight of the mass, in order to satisfy the condition of static equilibrium. Hence, 4FMg= 0. We then solve this equation forF.
Answer:F=Mg/4
