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danago
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A temporary mooring is to be set up on a concrete jetty. The mooring rope will exert a force (along the line of the rope) of 100 kN, and the maximum angle to the rope from the horizontal is 20 degrees. Specify the minimum block mass needed to safely secure the rope. Take the static coefficient of friction between concrete and concrete as \mu_{s}=0.5 and use a safety factor of 0.6. Only consider the slipping of the block.
I started by drawing a free body diagram of the situation and breaking the forces into components parallel and perpendicular to the ships surface.
I began by finding the required frictional force to balance the pull from the rope.
<br /> \sum {F_x = 0 \Rightarrow F_f - 100\cos (20) = 0} \therefore F_f = 100\cos (20) \approx 93.97kN<br />
Therefore, if the rope is pulling with 100kN, a frictional force of atleast ~94kN must be supplied to prevent slipping. Applying the safety factor, we get a force of:
<br /> f = \frac{{100\cos (20)}}{{0.6}} \approx 156.615kN<br />
The maximum static frictional force that can exist before the block starts slipping is given by:
<br /> f_{\max } = \mu _s \overline N = 0.5\overline N <br />
Where \overline N is the normal force provided by the ships surface. We want to find. The normal force required to meet the minimum requirements is given by:
<br /> 0.5\overline N = \frac{{100\cos (20)}}{{0.6}} \therefore \overline N = \frac{{100\cos (20)}}{{0.3}} \approx 313.23kN<br />
By summing the vertical component of all forces and knowing that the sum must equal zero, we can calculate the minimum weight force needed:
<br /> \begin{array}{l}<br /> \sum {F_y = 0} \Rightarrow 100\sin (20) + \overline N - W = 0 \\ <br /> \therefore W = 100\sin (20) + \overline N \approx 347.432kN \\ <br /> \end{array}<br />
Which corrosponds to a mass of ~35416kg.
Does that all look correct?
I started by drawing a free body diagram of the situation and breaking the forces into components parallel and perpendicular to the ships surface.
I began by finding the required frictional force to balance the pull from the rope.
<br /> \sum {F_x = 0 \Rightarrow F_f - 100\cos (20) = 0} \therefore F_f = 100\cos (20) \approx 93.97kN<br />
Therefore, if the rope is pulling with 100kN, a frictional force of atleast ~94kN must be supplied to prevent slipping. Applying the safety factor, we get a force of:
<br /> f = \frac{{100\cos (20)}}{{0.6}} \approx 156.615kN<br />
The maximum static frictional force that can exist before the block starts slipping is given by:
<br /> f_{\max } = \mu _s \overline N = 0.5\overline N <br />
Where \overline N is the normal force provided by the ships surface. We want to find. The normal force required to meet the minimum requirements is given by:
<br /> 0.5\overline N = \frac{{100\cos (20)}}{{0.6}} \therefore \overline N = \frac{{100\cos (20)}}{{0.3}} \approx 313.23kN<br />
By summing the vertical component of all forces and knowing that the sum must equal zero, we can calculate the minimum weight force needed:
<br /> \begin{array}{l}<br /> \sum {F_y = 0} \Rightarrow 100\sin (20) + \overline N - W = 0 \\ <br /> \therefore W = 100\sin (20) + \overline N \approx 347.432kN \\ <br /> \end{array}<br />
Which corrosponds to a mass of ~35416kg.
Does that all look correct?
