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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, m1 of mass 1.2 kg and m2 of mass 1.9 kg, are attached as shown to the opposite ends of the string, which passes over the pulley. The object m2 is in contact with the floor. (a) What is the largest value the force may have so that m2 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 m1?
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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