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.

Answer 1

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

$\sum{F_{y}}=0$

$N-W=0$

$N=W$

Put the value into the formula

$N=75\ lb$

Using balance equilibrium equation in horizontal

$\sum{F_{x}}=0$

$T_{2}-f_{k}=0$

$T_{2}=\mu_{k}N$

Put the value into the formula

$T_{2}=0.4\times75$

$T_{2}=30\ lb$

For pulley,

We need to calculate the force

Using balance equilibrium equation in horizontal

$\sum{F_{x}}=0$

$T\cos\theta+T\cos\theta=T_{2}$

$2T\cos30=T_{2}$

$T=\dfrac{T_{2}}{2\cos30}$

Put the value into the formula

$T=\dfrac{30}{2\times\cos30}$

$T=17.32\ lb$

We need to calculate the out put power

Using formula of power

$P=Tv$

Put the value into the formula

$P=17.32\times6$

$P=103.92\ lb.ft/s$

$P=0.188\ hp$

Hence, The out put power is 0.188 hp.

Answer 2

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 $\mu k$ = 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 $\theta = 30^{o}$.

Explanation:

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

           $\sum F_{y} = 0$

         N - W = 0

           N = W

or,      N = 75 lb

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

        $\sum F_{x} = 0$

       $T_{2} - F_{k}$ = 0

         $T_{2} = \mu_{k} N$

                  = $0.4 \times 75 lb$

                  = 30 lb

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

         $\sum F_{x} = 0$

       $T Cos (30^{o}) + T Cos (30^{o}) = T_{2}$

          $2T Cos (30^{o}) = T_{2}$

    or,             T = $\frac{T_{2}}{2 Cos (30^{o})}$

                        = $\frac{30}{2 Cos (30^{o})}$

                        = $\frac{30}{1.732}$

                        = 17.32 lb

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

             P = Tv

                = $17.32 lb \times 6$

                = $103.92 \times \frac{1}{550} \times \frac{hp}{ft/s}$

                = 0.189 hp

or,             = 0.2 hp

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