I need help on 11.98 c from University Physics 13th edition (Young). Knocking Ov

Question

I need help on 11.98 c from University Physics 13th edition (Young).

Knocking Over a Post. One end of a post weighing 400 N and with height h rests on a rough horizontal surface with The upper end is held by a rope fastened to the surface and making an angle of with the post (Fig. P11.98). A horizontal force is exerted on the post as shown. (a) If the force is applied at the midpoint of the post, what is the largest value it can have without causing the post to slip? (b) How large can the force be without causing the post to slip if its point of application is of the way from the ground to the top of the post? (c) Show that if the point of application of the force is too high, the post cannot be made to slip, no matter how great the force. Find the critical height for the point of application.

Explanation / Answer

There are 3 vertical forces. The vertical component of the tension in the rope, the weight of the post, and the force exerted by the surface. The force exerted by the surface in the opposite direction of the vertical component of tension in the rope and the weight of the post.
The vertical component of tension in the rope = T * cos 36.98
Surface force = T * cos 36.98 + 400

There are 3 horizontal forces. The force, the horizontal component of tension in the rope, and the friction force. The force is in the opposite direction of the horizontal component of tension in the rope and the friction force
The horizontal component of tension in the rope = T * sin 36.98
Friction force = µ * surface force = 0.30 * (T * cos 36.98 + 400)
F = (T * sin 36.98) + [0.30 * (T * cos 36.98 + 400)]

As the force is applied, the bottom of the post tends to slide in the direction of the force. This causes the post to rotate.
Let the pivot point be at the middle of the post.
Clockwise torque = T * sin 36.98 * h/2
Counter clockwise torque = Ff * h/2 = 0.30 * (T * cos 36.98 + 400) * h/2
Clockwise torque = Counter clockwise torque
T * sin 36.98 * h/2 = 0.30 * (T * cos 36.98 + 400) * h/2
Divide both sides by h/2
T * sin 36.98 = 0.30 * (T * cos 36.98 + 400)
T * sin 36.98 = (0.30 * T * cos 36.98) + (0.30 * 400)
T * sin 36.98 – (0.30 * T * cos 36.98) = (0.30 * 400)
T * (sin 36.98 – 0.30 * cos 36.98) = (0.30 * 400)
T = (0.30 * 400) ÷ (sin 36.98 – 0.30 * cos 36.98)
T = 331. 6 N
Substitute 331.6 for T in the equation below.
F = (T * sin 36.98) + [0.30 * (T * cos 36.98 + 400)]


(Part B)
How large can the force be without causing the post to slip if its point of application is 6/10 of the way from the ground to the top of the post?
Now the distance from the friction force = 0.6 * h
Distance from horizontal component of tension in the rope to F = 0.4 * h

Clockwise torque = T * sin 36.98 * 0.4 * h
Counter clockwise torque = 0.30 * (T * cos 36.98 + 400) * 0.6 * h
T * sin 36.98 * 0.4 * h = 0.30 * (T * cos 36.98 + 400) * 0.6 * h
T * sin 36.98 * 0.4 * h = 0.18 * (T * cos 36.98 + 400) * h

Divide both sides by 0.4 * h
T * sin 36.98 = 0.45 * (T * cos 36.98 + 400)

Solve for T as before. And then solve for F

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