Question

MR 772 A block of mass m2 = 1.08 kg and a block of mass m2 = 12.1 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 = 2 kg. The fixed, wedge-shaped ramp makes an angle of 0 = 30.0° as shown in the figure. The coefficient of kinetic friction is 0.19 for both blocks. Determine the acceleration of the blocks.

Answer 1

Solution : Free body diagram of all the components is : ANO = mig T mi fr = u No v mig where, Ni = Normal reaction = mig (weight) (3 EF,-0) a = assumed acceleration of blocks in right direction T, = tension due to string fi - kinetic friction as block is moving towards right with acceleration'a'.. EMEN, de EN, : Defination or fi = Mx mig. -11). M₂ : R f=MxN2 KIV m2 g a ima maguso mag sino where, N2 = normal reaction a = acceleration of block consistent with our assumption while making free body diagram of mass my T2 = Tension due to string. f2 = kinetic friction or for or, for My Nz = Mx Mag coso. (as N₂ = m_guose a because block is not leaving the surface of ramp).

- 3) E Fxoma mia. ( Newton's second low) =) T, - Up No = mia. Ti - Me mig =m, a. (: equation s)) » Tizm, a + Mx mig. < F along ramp, me = m2a. mag sino - Tz - f = mza. ► mağ sino – T2 – Uk meg.cos @ = maa (: » T2 = mag sino - Ukm₂ g cose -mza - (4) pulley Tis d a= a (pure rolling) C equation 'second Law ) it's axis Torque of pulley: (T2-T, DR = IQ (Newtons =(T2-T) R' = MRZ as, moment of inertia , I - MR? for disc (pulley) about Ž and a = a. > [m29 sine - Mk mag cos o - m, a)-imat Axmg)] - Ma (i equation 18 ) and (17) - [m2 (sino - Urcose) – milk ]9 -15). M + mi + m2 a acceleration due to gravity Putting values in (5), we get, 36:=9.81 m) a = 2.66 m

Draw the Free body diagram of each of the component. Write equations according to Newton's Second Law of motion.

The blocks are not making any motion in perpendicular direction of the ramp, Hence, the sum of forces about that direction is zero.

The acceleration of the blocks is 2.66 m/s^2.