The block has a weight of 75 lblb and rests on the floor for which μkμk = 0.4. The motor draws in the cable at a constant rate of 6 ft/sft/s

Question

The block has a weight of 75 lblb and rests on the floor for which μkμk = 0.4. The motor draws in the cable at a constant rate of 6 ft/sft/s. Neglect the mass of the cable and pulleys.

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Orla Orla 1 year 2021-09-03T04:04:29+00:00 2 Answers 8 views 0

Answers ( )

    0
    2021-09-03T04:05:41+00:00

    Answer:

    The out put power is 0.188 hp.

    Explanation:

    Given that,

    Weight = 75 lb

    Coefficient of friction = 0.4

    Rate = 6 ft/s

    Suppose, Determine the output of the motor at the instant θ = 30°.

    For block,

    We need to calculate the force in vertical direction

    Using balance equilibrium equation in vertical

    [tex]\sum{F_{y}}=0[/tex]

    [tex]N-W=0[/tex]

    [tex]N=W[/tex]

    Put the value into the formula

    [tex]N=75\ lb[/tex]

    Using balance equilibrium equation in horizontal

    [tex]\sum{F_{x}}=0[/tex]

    [tex]T_{2}-f_{k}=0[/tex]

    [tex]T_{2}=\mu_{k}N[/tex]

    Put the value into the formula

    [tex]T_{2}=0.4\times75[/tex]

    [tex]T_{2}=30\ lb[/tex]

    For pulley,

    We need to calculate the force

    Using balance equilibrium equation in horizontal

    [tex]\sum{F_{x}}=0[/tex]

    [tex]T\cos\theta+T\cos\theta=T_{2}[/tex]

    [tex]2T\cos30=T_{2}[/tex]

    [tex]T=\dfrac{T_{2}}{2\cos30}[/tex]

    Put the value into the formula

    [tex]T=\dfrac{30}{2\times\cos30}[/tex]

    [tex]T=17.32\ lb[/tex]

    We need to calculate the out put power

    Using formula of power

    [tex]P=Tv[/tex]

    Put the value into the formula

    [tex]P=17.32\times6[/tex]

    [tex]P=103.92\ lb.ft/s[/tex]

    [tex]P=0.188\ hp[/tex]

    Hence, The out put power is 0.188 hp.

    0
    2021-09-03T04:05:47+00:00

    The given question is incomplete. The complete question is as follows.

    The block has a weight of 75 lb and rests on the floor for which [tex]\mu k[/tex] = 0.4. The motor draws in the cable at a constant rate of 6 ft/sft/s. Neglect the mass of the cable and pulleys.

    Determine the output of the motor at the instant [tex]\theta = 30^{o}[/tex].

    Explanation:

    We will consider that equilibrium condition in vertical direction is as follows.

               [tex]\sum F_{y} = 0[/tex]

             N – W = 0

               N = W

    or,      N = 75 lb

    Again, equilibrium condition in the vertical direction is  as follows.

            [tex]\sum F_{x} = 0[/tex]

           [tex]T_{2} – F_{k}[/tex] = 0

             [tex]T_{2} = \mu_{k} N[/tex]

                      = [tex]0.4 \times 75 lb[/tex]

                      = 30 lb

    Now, the equilibrium equation in the horizontal direction is as follows.

             [tex]\sum F_{x} = 0[/tex]

           [tex]T Cos (30^{o}) + T Cos (30^{o}) = T_{2}[/tex]

              [tex]2T Cos (30^{o}) = T_{2}[/tex]

        or,             T = [tex]\frac{T_{2}}{2 Cos (30^{o})}[/tex]

                            = [tex]\frac{30}{2 Cos (30^{o})}[/tex]

                            = [tex]\frac{30}{1.732}[/tex]

                            = 17.32 lb

    Now, we will calculate the output power of the motor as follows.

                 P = Tv

                    = [tex]17.32 lb \times 6[/tex]

                    = [tex]103.92 \times \frac{1}{550} \times \frac{hp}{ft/s}[/tex]

                    = 0.189 hp

    or,             = 0.2 hp

    Thus, we can conclude that output of the given motor is 0.2 hp.

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