Question #143668
A block of mass m1 = 2.00 kg and a block of mass m2 = 6.00 kg are connected by a massless string over a pulley in the shape of a solid disk having radius R = 0.250 m and mass M = 10.0 kg. The fixed, wedge- shaped ramp makes an angle of 휃 = 30° as shown in the figure. The coefficient of kinetic friction is 0.360 for both blocks. (a) Draw force diagrams of both blocks and of the pulley. Determine (b) the acceleration of the two blocks and (c) the tensions in the string on both sides of the pulley.
1
Expert's answer
2020-11-12T09:35:46-0500

(a) The diagram:



(b) To find the acceleration, use Newton's second law (regular one with forces and the one for case of rotation). Assume positive x is to the right and y is to upward.


  • Forces acting on m1:
T1f1=m1a,N1m1g=0.T1=µm1g+m1a.T_1 − f_1 = m_1a,\\ N_1-m_1g=0.\\ T_1 = µm_1g + m_1a.



  • Forces acting on m2:
T2f2+m2g sinθ=m2a,N2m2g cosθ=0.T2=m2g sinθµm2g cosθm2a−T_2 − f_2 + m_2g \text{ sin}θ = m_2a,\\ N_2-m_2g\text{ cos}\theta=0.\\ T_2 = m_2g \text{ sin}θ − µm_2g \text{ cos}θ − m_2a

  • Torque on the pulley:
τnet=Iα,T2RT1R=IaR, a=2R2MR2(T2T1).τ_\text{net} = Iα,\\ T_2R − T_1R = I\frac aR,\\\space\\ a =\frac{2R^2}{MR^2}(T_2 − T_1).

Express the acceleration:

a=2g[m2 sinθµ(m2 cosθ+m1)]M+2(m1+m2)=0.309 m/s2.a=\frac{2g[m_2 \text{ sin} θ − µ(m_2 \text{ cos} θ +m_1)]}{M + 2(m_1 + m_2)}=0.309\text{ m/s}^2.


(c) Using the equations for the objects of the system in part (b), find the tensions:


T1=m1(µg+a)=7.674 N,T2=m2(g sinθµg cosθa)=9.21 N.T_1 = m_1(µg + a)=7.674\text{ N},\\ T_2 = m_2(g \text{ sin} θ − µg\text{ cos} θ − a)=9.21\text{ N}.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS