Respuesta :
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.
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.
