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

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.

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.