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Someone exerts a force F directly up on the axle of the pulley shown in Fig. Consider the pulley and string to be mass- less and the bearing frictionless. Two objects, m 1 of mass 1.2 kg and m 2 of mass 1.9 kg, are attached as shown to the opposite ends of the string, which passes over the pulley. The object m 2 is in contact with the floor. (a) What is the largest value the force may have so that m 2 will remain at rest on the floor? (b) What is the tension in the string if the upward force F is 110 N? (c) With the tension determined in part (b), what is the acceleration of m 1 ?
Round off to two significant figures,F = 37 NTherefore, the maximum force F for which block with mass m2 stays on floor is 37 N .(b)The tension in the string is related to upward force as:Therefore, the tension in the string corresponding to upward force of 110 N is 55 N.
a. Taking the same situation mid air we get that m2 goes down with the accelaration of a = (m2-m1)g/(m2+m1). Now considering this case for the mass m2 to remain at the ground the pulley shold go up 2a. so F=(m1+m2)2a = 2(m2-m1)g. putting values we get the answer = 14N pls aprove if it is helpful to you
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