Some miners wish to remove water from a mine shaft. A pipe is lowered to the water 90 m below, and a negative pressure is applie
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Answer:
a=g(sinθ-μkcosθ)
Explanation:
In an inclined plane the forces that interact with the object can be seen in the figure. The normal force, the weight w and the decomposition of the force vector of weight can be observed.
wx=m*g*sinθ
wy=m*g*cosθ
As the objects moves down an incline, acceleration in y axis is 0.
Then, by second Newton's Law:
Fy = m*ay
FN - m*g cos θ = 0,
FN=m*g cos θ
In x axis the forces that interacs are the x component of weight and friction force:
Fx = m*ax
mg sen u-FN*μk=m*a
Being friction force, Fr=FN*μk, we replace with its value in below formula:
m*g *sinθ-(m*g*cosθ*μk)=m*a
Then, isolating a:
a=(m*g sinθ-(m*g*cosθ*μk))/m
Solving, we have next equation:
a=g sinθ-(g*cosθ*μk)
Applying distributive property we have:
a=g*(sinθ-μk*cosθ)
The two possible angles obtained by using the qudratic equation are;
θ = 15.10° and θ2 = 73.51°
Given, speed of water = = 50ft/s
For the motion along x direction, time period can be calculated as follows:
35 = (50 × cosθ) t
t = 0.64 / cosθ
For the motion in y direction, an equation can be obtained as follows:
θ)
Plugging in the values we get:
θ)
-20 = -32tanθ - 10.304θ
Upon solving the above quadratic equation, we get,
tanθ = 0.27 , -3.38
Therefore,
tanθ = 0.27
θ = 15.10°
and, tanθ = -3.38
θ = 73.51
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