Question

Determine the angle φ at which the applied force P should act on the pipe so that the magnitude of P is as small as possible for pulling the pipe up the incline. What is the corresponding value of P? The pipe weighs W and the slope α is known. Express the answer in terms of the angle of static friction, θ = tan-1 μs.

Answer 1

Answer:

∅=Ф  

P = W sin($\alpha$ + Ф)

Explanation:

First, we'll isolate and draw the free-body diagram of the pipe  

Note that since the pipe is moving, the friction force is equal to the product of normal reaction force and the kinetic coefficient of friction  

F = F_max = u_kN  

Also note that the weight makes with the y-axis angle a because the x-axis makes the same angle with the horizontal  

The expression for angle of friction is:

B = tan-1 (u_k)

From here we can express the coefficient of friction as:

u_k = tan(Ф)

Replace u_s by tan(Ф) in the expression for the friction force

F = N tan(Ф)  

diagram is attached

By equating sum of forces in y-direction to zero, we can write the expression for the normal reaction force  

ΣF_y = 0

N — W cos$\alpha$- P sin Ф= 0

From here we can express N as:

N = W cos$\alpha$ -— P sin Ф

Replace N by the expression above in the expression for friction force F(written in step 1)  

F = (W cos$\alpha$  — P sin  Ф) tan( Ф)                                 (1)  

Now, we'll equate sum of forces in x-direction to zero  

ΣF_x = 0

-F - W cos$\alpha$  + P sin  Ф =0

Replace F by expression (1)  

— (W cos$\alpha$  — P sin Ф) tan(Ф) — W sin$\alpha$+pcosФ=0

-W cos $\alpha$ tan(Ф) + P sin Ф tan(Ф) — W sin$\alpha$ +pcosФ=0

P(sin Ф tan(Ф) + cosФ) — W(cos $\alpha$ tan(Ф) + sin $\alpha$)

From here we can express the force P needed to pull the pipe as:

P = W(cos$\alpha$  tan(Ф) + sin$\alpha$)/sinФ*tansФ+cosФ                    (2)

All we have to do now is to simplify the expression (2). We'll start by sin replacing tan(Ф) with sinФ/cosФ

P = W(cos *sinФ/cosФ + sin)/sinФ*sinФ/cosФ+cosФ *cosФ/cosФ

We can multiply the right side of equation by cosФ/cosФ

P = W(cos$\alpha$ *sinФ + sin$\alpha$cosФ)/sin∅*sinФ+cos∅cosФ *cosФ/cosФ

Finally, we'll replace (cos$\alpha$ *sinФ + sin$\alpha$cosФ) by sin($\alpha$ + Ф) and (sin∅ sinФ + cos∅ cos Ф) by cos( ∅— Ф)

P wsin($\alpha$ + Ф) /cos(∅ — Ф)                                              (3)  

Since the first derivative of the function is actually tangens of the angle which tangent makes with the x-axis, we'll find it by equating the first derivative by zero(this means that the tangent of the function is horizontal, i.e. that the function is at its maximum or minimum)  

Note that the variable in the expression (3) is 0, since both B and a are known  

dP/d∅ =d/d∅ [sin(Ф+)/cos(∅-Ф) ]

Note that sin(Ф+$\alpha$) is constant since both Ф and a are known  

dP/d∅ = sin(Ф+$\alpha$) d/dФ [1/cos(∅-Ф) ]  

Next, we'll apply the reciprocal rule  

= -dP/d∅[cos(∅-Ф)]/cos^2(∅-Ф)*sin(Ф+$\alpha$)

Next, we'll apply the differentiation rule  

=(-sin(∅-Ф))*d/d∅[∅-Ф]*sin(Ф+$\alpha$)/cos^2(∅-Ф)

=(d/d∅[∅]+d/d∅[-∅])*sin(Ф+$\alpha$)sin(∅-Ф)/cos^2(∅-Ф)

dP/d∅ =sin(Ф+$\alpha$)*sin(∅-Ф)/cos^2(∅-Ф)                       (4)

Next step will be to equate the expression (4) to zero, to determine the value of # when the function is minimum  

sin(Ф+$\alpha$)*sin(∅-Ф)/cos^2(∅-Ф) =0  

Note that sin(Ф+$\alpha$) is constant, so in order for the equation above to be correct, sin(∅-Ф) needs to be equal to zero  

sin(∅-Ф)  = 0

Since sin 0° = sin 180° = 0, two possible solutions for ∅ are:

∅-Ф=0                           Ф=∅  

or  

∅-Ф = 180°                    ∅ = 180° +  Ф

Since the function for P is only good over the range 0 <  ∅ < 90°, since when > 90° the friction force will change its direction, we can conclude that the minimum force P is required to move the pipe at angle:  

∅=Ф  

Finally, replace # by 8 in expression (3) to determine the minimum force P required to move the pipe

P = W sin($\alpha$ + Ф ) / cos ∅ —  ∅)  

P = W sin($\alpha$ + Ф)