Skip to main content

When the blocks are released, what is the acceleration of the lighter block?

A 1.4 kg block and a 2.7 kg block are attached to opposite ends of a light rope. The rope hangs over a solid, frictionless pulley that is 25 cm in diameter and has a mass of 0.72 kg. When the blocks are released, what is the acceleration of the lighter block?I think I’ve figured out that the force equation for the 1.4 kg block is T1-m1g=m1a, and for the 2.7 kg block: T2-m2g=m2a. I believe the torque equation for the pulley is torque=rF, where F would be (T2+T1), I think. And that is set equal to I*alpha, or (mr^2)*(a/r). But, I really have no idea what to do beyond that. I tried solving for a by rearranging the force equations to equal T1 and T2, and then plugged in those values into T1 and T2 in the torque equation, but my answer was wrong. Any help or clarification would be greatly appreciated, and will get the best answer.

2 Answers

m₁ = mass 1 = 1.4 kg m₂ = mass 2 = 2.7 kgm₃ = mass of the pulley = 0.72 kgR = radius of the pulley = 0.25 mg = accelerayion by gravity = 9.8 m/s²First, you need to ‘choose’ a direction in which the system will accelerate. As m₂ is bigger than m₁, it seems logic to choose m₂ moving down, and m₁ moving up.(Note: If you choose wrong, that’s not a problem. You will end up with a negative value for acceleration so you will know that the system moves in the other direction than the one you chose)The forces acting on m₁ are:* tensile force in the rope, pulling up * force of gravity, pulling downAs m₁ is moving up, you know that the tensile force is bigger than the gravitational force. The net force on m₁ will be:Fnet₁ = T₁ – m₁×gBy Newton’s second law, you can say:T₁ – m₁×g = m₁×aT₁ = m₁×g + m₁×a . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . equation 1The forces acting on m₂ are:* tensile force in the rope, pulling up* force of gravity, pulling downAs m₂ is moving down, you know that the gravitational force is bigger than the tensile force. The net force on m₂ will be:Fnet₂ = m₂×g – T₂By Newton’s second law, you can say:m₂×g – T₂ = m₂×aT₂ = m₂×g – m₂×a . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . equation 2The moment of inertia of the pulley (assuming it is a solid disk) is:I = m₃×R²/2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . equation 3The equation for torque on the pulley is:T₂×R – T₁×R = I×a/RRearranging this a bit, you get:(T₂ – T₁)×R = I×a/RT₂ – T₁ = I×a/R² . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . equation 4 Substituting equation 3 into equation 4, you get:T₂ – T₁ = (m₃×R²/2)×a/R²T₂ – T₁ = m₃×a/2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . equation 5(note how the radius R cancels out)Now substituting equations 1 and 2 into equation 5, you get:(m₂×g – m₂×a) – (m₁×g + m₁×a) = m₃×a/2m₂×g – m₂×a – m₁×g – m₁×a = m₃×a/2m₂×g – m₁×g = m₃×a/2 + m₂×a + m₁×a(m₂ – m₁)×g = (m₃/2 + m₂ + m₁)×aa = g × (m₂ – m₁) / (m₃/2 + m₂ + m₁)And finally, entering the numbers, you get:a = (9.8 m/s²) × [(2.7 kg) – (1.4 kg)] / [(0.72 kg)/2 + (2.7 kg) + (1.4 kg)]a = 2.8565… m/s²a = 2.9 m/s² (to 2 sf)... Show More
(T2+T1)… This is causing the issue.for 1.4 Kg block I have F2=m2g so we have T2 – m2g = mafor 2.7 Kg block I have F1=m1g and m1g – T1 = maThe torque is caused F1=m1g because its has more mass which caused the other block to accelerate upward.... Show More

Want to answer this question?

Join our community to share your knowledge!

Categories

Tags

Related Questions

10
When the dollar ‘rises’ compared to other currencies, this means

When the dollar "rises" compared to other currencies, this means

4
When the linear form of glucose cyclizes, the product is a(n): Group of answer choices Anhydride Glycoside Hemiacetal Oligosaccharide

When the linear form of glucose cyclizes, the product is a(n): Group of answer choices Anhydride Glycoside Hemiacetal Ol...

10
When the monster and frankenstein finally meet again, how are they different?

A; Frankenstein is tortured by guilt, whereas the monster is tortured by rejection - APEX!!

4
when they ask “subjects studied” on job application?

I am filling out a job application. In the section where they ask you the name of your high school, there is also a spac...

2
When to use GST Collected and GST Credits Received?

I’m a year 11 student studying accounting. I have been understanding everything we’ve learned so far, but now that we’re...

6
When using the note-card system, what goes in the upper right-hand corner of the card?

When using the note-card system, what goes in the upper right-hand corner of the card?

10
When using the perfect tenses of verbs, it’s necessary to add a/an

When using the perfect tenses of verbs, it's necessary to add a/an a. irregular verb. b. infinitive. c. verb. d. suffix.

2
When using the periodic system the physical inventory count is used to determine?

When using the periodic system the physical inventory count is used to determinea) only the sales value of goods in the...