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M1 - moments and beams
This is the only topic of M1 that I still have problems with... so can someone describe to me how exactly I should go about solving these questions?
A uniform beam AB has mass 12 kg and length 3 m. The beam rests in equilibrium in a horizontal position, resting on two smooth supports. One support is at end A, the other at a point C on the beam, where BC = 1 m, as shown in Figure 3. The beam is modelled as a uniform rod.
(a) Find the reaction on the beam at C.
A woman of mass 48 kg stands on the beam at the point D. The beam remains in equilibrium.
The reactions on the beam at A and C are now equal.
(b) Find the distance AD.
Any help is seriously appreciated...
A uniform beam AB has mass 12 kg and length 3 m. The beam rests in equilibrium in a horizontal position, resting on two smooth supports. One support is at end A, the other at a point C on the beam, where BC = 1 m, as shown in Figure 3. The beam is modelled as a uniform rod.
(a) Find the reaction on the beam at C.
A woman of mass 48 kg stands on the beam at the point D. The beam remains in equilibrium.
The reactions on the beam at A and C are now equal.
(b) Find the distance AD.
Any help is seriously appreciated...
Right, well, two things you normally need to do in this type of problem:
1. The rod is uniform, so its centre of mass is in the middle. Take moments about a convenient point (here it looks convenient to take moments about A, because if we take moments about A we don't have to consider the reaction force at A, and we'll just get something involving the reaction at C and the weight).
2. Resolve vertically. The sum of the upward reaction forces is the same as the weight of the rod.
1. The rod is uniform, so its centre of mass is in the middle. Take moments about a convenient point (here it looks convenient to take moments about A, because if we take moments about A we don't have to consider the reaction force at A, and we'll just get something involving the reaction at C and the weight).
2. Resolve vertically. The sum of the upward reaction forces is the same as the weight of the rod.
Why is it inconvenient to take moments about, say, C?
EvenStevens
Why is it inconvenient to take moments about, say, C?
It's not "inconvenient", but you want to find the reaction at C. Remember that moment is force * distance, so if you take moments about C, you won't take into account the force at C (and so you're not really much closer to finding out what that force is). It's perfectly ok to take moments about C, find the reaction at A and then resolve forces to find the reaction at C - it'd still work. Mechanics is often a case of just "doing things" until the answer you want comes out. But if you can spot a short-cut, e.g. taking moments about A here, you might as well take it.
this question is very simple to complete
basic moments
with practice you will be able to do this in a breeze
just follow what generalebriety said and your fine
i did this in about 3mins =]
basic moments
with practice you will be able to do this in a breeze
just follow what generalebriety said and your fine
i did this in about 3mins =]
Aaah, I see.
Thanks a bunch :-)
Thanks a bunch :-)
No problem. 
One more question
For part b, where a woman of mass 48kg stands on the beam, is there a normal reaction on the woman or not?
For part b, where a woman of mass 48kg stands on the beam, is there a normal reaction on the woman or not?
Original post by vb789
Dont think it really matters but D'll probably be between A and C, closer to C.
have a look at this then, i put D close to C and tried working it out but my answer is diiferent from the answer in the book.
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I got part a right a right but i dont understand where i am going wrong part b. any idea?
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