Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
KrisConnett
Group Title
A mass of 1kg on an incline of 37 degrees is being pulled with a force of 12 N upward that incline. The mass is attached to a string going to the bottom of the incline, which is attached to a pulley, which is attached to a string attached to a mass of 3kg on the ground. Assuming there is no friction and the mass of the string and pulley are negligible, what is the tension of the string?
 2 years ago
 2 years ago
KrisConnett Group Title
A mass of 1kg on an incline of 37 degrees is being pulled with a force of 12 N upward that incline. The mass is attached to a string going to the bottom of the incline, which is attached to a pulley, which is attached to a string attached to a mass of 3kg on the ground. Assuming there is no friction and the mass of the string and pulley are negligible, what is the tension of the string?
 2 years ago
 2 years ago

This Question is Open

KrisConnett Group TitleBest ResponseYou've already chosen the best response.0
dw:1349917367041:dw
 2 years ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
dw:1350795258686:dw
 2 years ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
dw:1350795463231:dw
 2 years ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
For 3kg block \[T = m_1a \space \space \space \space \space \space ..........(i)eq.\]\[m_1=3kg.\]We would ignore Normal reaction & its weight because they balance each other. So no acceleration in ydirection. But there is acceleration in xdirection so we have made above equation.
 2 years ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
Here is Force diagram of 1kg block dw:1350796181544:dw
 2 years ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
here also normal & m2g cos theta would balance each other. Therefore no need of them. we need only xdirection forces because acceleration is only in xdirection. So our next equation is \[12(T+m_2g \sin \theta)= m_2a \space \space \space \space \space ..........(ii)eq.\]\[m_2=1kg.\]
 2 years ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
Now u know all the values because they are given in the question itself. So solve eq. (i) & (ii). U will get T which is the tension in the rope. But \[\LARGE{\color{red}{\text{Please read slowly all the stuff.}}}\] Thanx:)
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.