Question

In the system shown in Figure P5.31, a horizontal force F acts on the 9.50 kg object (). The horizontal surface is frictionless. Figure P531 (a) For what values of x does the 4.50 kg ohject (m2) accelerate upward? (b) For what values of Fy is the tension in the cord zero?

Answer 1

Solution)

Let tension be T1

So,

Fx - T = m2 a ;

Fx - T = 9.5a ; eq (1)

and T - 4.5 * g = 4.5 * a ;

T = 4.5 ( a + g ) ;

putting values in 1 ;

Fx - 4.5 ( a + g ) = 9.5 a

Fx = 4.5 ( a + g ) + 9.5 a

Here, a = 0

Fx = 4.5 * 9.8 = 44.1

so Fx > 44.1 N (Ans)

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( b) the direction is negative (towardsright )

Here,T = 0 ,

T - 4.5 * g = 4.5 * a

Therefore, a = g ;

and Fx - T = m2 *a

Fx = m2 * g = 9.5 * 9.8 = 93.1 N in the negative x dirtection ,

Fx < = - 93.1 N (Ans)

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Good luck!:)