## A statics problem

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Darkstorm
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### A statics problem

I'm struggling with finding the solution to this problem:

A horizontal, even bar (AB) has these properties:
Length= 4 meters
Weigth= 75 kg

The beam is supported at A and C with BC being 0.8 meters

Calculate the vertical forces on the beam at A and C

With 18.75 kg/m, BC weighs 15kg and AC weighs 60 kg

Gravity total = ~736 Newton
Gravity BC = ~148 Newton
Gravity AC = ~588 Newton

I just can't wrap my head around how to calculate the forces in A and C
Any help is appreciated.

measure
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Location: Time-traveling kayak

### Re: A statics problem

There are two unknowns in the problem (the reactions at A and B), so you'll need two independent equations to solve. The first will be application of Newton's 2nd law to the linear motion of the beam: sum of forces equals mass times acceleration (which is zero). The second equation will again be from Newton's 2nd law, but applied to the angular motion: sum of torques equals moment of inertia times angular acceleration (also zero).

Darkstorm
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Location: Sirdal, Norway
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### Re: A statics problem

measure wrote:There are two unknowns in the problem (the reactions at A and B), so you'll need two independent equations to solve. The first will be application of Newton's 2nd law to the linear motion of the beam: sum of forces equals mass times acceleration (which is zero). The second equation will again be from Newton's 2nd law, but applied to the angular motion: sum of torques equals moment of inertia times angular acceleration (also zero).

Yeah, I know that these equations should add up to zero, but my trouble is finding how much normal force is applied to the beam at A and C

measure
Posts: 126
Joined: Sat Apr 04, 2015 4:31 pm UTC
Location: Time-traveling kayak

### Re: A statics problem

Darkstorm wrote:Yeah, I know that these equations should add up to zero, but my trouble is finding how much normal force is applied to the beam at A and C

Start by setting up the two equations: What are the forces in the vertical direction? There is the weight of the beam pulling down, and the two normals pushing up. Since the beam doesn't accelerate, these must balance, and so the sum of the normals equals the total weight of the beam. From the second equation, you will be able to find the difference between the two normal forces. Once the sum and difference are known, the rest is substitution and algebra.

Alternatively, if you consider the moments acting about one of the reaction points (say, point A), since the normal at A has a zero-length lever arm, it has no contribution to the moment. Then the moment equation has only a single unknown, the normal at B, which you can solve for.

drachefly
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Joined: Thu Apr 23, 2009 3:25 pm UTC

### Re: A statics problem

I would suggest calculating the torque around the center of mass of the beam - the force of gravity will then exert zero torque, and you can focus on the forces you care about more.

44 stone lions
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Joined: Fri Jan 07, 2011 5:57 pm UTC

### Re: A statics problem

But calculating the torque or moment about the centre of mass of the beam gets you an expression with two unknowns, and you can only solve for one. That's why you take the moment about one of the reaction points, since there's no distance the force exerted on the beam produces no torque at that point. Since you know the other distances in involved plus the other vertical forces, with the exception of the other reaction point which becomes the single unknown in the equation which you can solve for.

At least that the way I remember it, it has been a while. Although my mechanics lecturer did once make us stand at the front of the class and do what I can only describe as some form of hula dance to help us remember the condition for equilibrium, I remember that quite vividly

drachefly
Posts: 194
Joined: Thu Apr 23, 2009 3:25 pm UTC

### Re: A statics problem

True - I forgot that the question was to get the force (doing that would get you one of the forces), not get the ratio of the two forces (doing it my way would yield that very directly).