Question

(10%) Problem 8: Two blocks connected by a string are pulled across a horizontal surface by a force applied to one of the blocks, as shown to the right. The mass of the left block m1 = 1.3 kg and the mass of the right block m2 = 3.1 kg. The angle between the string and the horizontal is = 61° The coefficient of kinetic friction between the blocks and the surface is u= 0.22. Each block has an acceleration of a=2.9 to the right. m m, 50% Part (a) Write an equation for the magnitude of the force exerted by the string. Use variables from the problem statement as well as using g for the gravitational acceleration Grade Summary F = cos(e) - u sin() Deductions 0% Potential 100% cos(a) cos(o) cos() ( ) 7 8 9 HONE Submissions Attempts remaining: 13 sin(a) sin(o) sin(e) 4 5 6 - (0% per attempt) detailed view 1 * 1 2 3 - 0% mi+ 0 . END VO BACKSPACE DU CLEAR Hint Submit Hints: 0 for a 0% deduction. Hints remaining: 0 F eedback I give up! Feedback: 0% deduction per feedback.

Answer 1

Given € Let the tension in the string be Drawing the each block free body diagrame Esino a COSCO Where, e & N f, & fe offered N are the frictional forces by m, 4 m₂ respectively Normal forces

for mass mi using and Law of Newton Almg x - axis Elnet - Mi ay T- f = max = m, a ax= acceleratsen along X-axissa ay = 0 Along y-axis, con Ni-mig = milay = 0 Ni migo N = mig frictional force We can write fi= M. Ni - umig um, q Equation ③ becomes T-u mig - mia T = umigt mia -

For mass m2 Using and law of Newton Along x-axis, E Free - Max FCOSO - T- ta = ma - Along y-axis , { thet = Melly = of Mz + ESINO — My = 0 N2 = Mg - Fsino Frictional force, f = el Na = u. (mag - F since ) eq. ③ becomes Froso - T - u mg - Fsince = Maa @

NA MINISTER ME T ERMINALABANWALSHOWAWALAMI Writing Putting the value of T from 6 ECOSO - mia - umg - sing at ut since - Maa F (Coscot usino) - (mith) at ulm+ M2), (mitm) at uforitme) g Teoso + using) (at up) (m+ma) (cose + usine ) F