What is the required horizontal force to deflect a hanging rod by 30 degrees?

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To deflect a uniform rod AB, hinged at point A, by 30 degrees from the vertical, a horizontal force at point B is required. The calculations involve taking moments about point A, leading to the equation Fcos30° x AB = wsin30° x AB/2, where F is the applied force and w is the weight of the rod. It is clarified that the weight's moment should use sin30° instead of cos30°. Additionally, the term "a gauge of 56.5 ins" refers to the inner distance between railway tracks. Accurate moment calculations are crucial for determining the required force.
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A uniform rod AB, of mass 5lb, is hinged at upper end A and hangs down freely. What horizontal force applied at B will be required in order to deflect the rod through 30 degrees from the vertical. Thanks.
My attempt: taking moments about A, Fcos30xAB = wxAC = wcos30xAB/2 therefore F = w/2. F= applied horizontal force, w = weight.A 2nd question what does "a gauge of 56.5 ins" mean in rhe context of a railway line?
 
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John O' Meara said:
My attempt: taking moments about A, Fcos30xAB = wxAC = wcos30xAB/2 therefore F = w/2. F= applied horizontal force, w = weight.A
Careful. Note that the weight acts vertically, so its moment should have a sin30, not a cos30.

2nd question what does "a gauge of 56.5 ins" mean in rhe context of a railway line?
The inner distance between the rails.
 
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I would take a second look and the Fcos30xAB. Shouldn't it be sin to get the correct length of the normal distance?
 
civil_dude said:
I would take a second look and the Fcos30xAB. Shouldn't it be sin to get the correct length of the normal distance?
Actually, that part is correct, since the force is applied horizontally. But the moment due to the weight should be Wsin30xAB/2, not Wcos30xAB/2. My bad, for not reading carefully. :redface: I'll revise my comments in the original post.
 
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Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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