SUPER Hard Physics Problem!!! Please Help I am Desperate!?

Question

A uniform beam has a mass of 236 kg and has 5 forces working on it. F1 and F2 are pushing upward at each end of the beam while F3 (3900 N), F4 (2990 N), and F5 (1960 N) are pushing downward in between the two upward forces.

There are four distances. D1 (1.84 m) is between F1 and F3. D2 (4.16 m) is between F3 and F4. D3 (2.90 m) is between F4 and F5. D4 (1.02 m) is between F5 and F2.

Calculate F1 and F2 for the beam.

Can anyone explain how to do this problem?? I have drawn a free-body diagram, but I am at a complete loss as to what to do now.

Answer 1

do it in 2 parts,
total upward forces= total downward forces F1+F2=F3+F4+F5+236*9.81
F1+F2=11165.16N
then,
closewise moments = anti clockwise moments
F1*4.96+3.94*1960+1.04*2990=F2*4.96+3.12*3900
4.96F1=4.96F2+1336
F1=F2+269.355

F1=5717.3N
F2=5447.9N

Answer 2

It must have been mentioned that the beam is in equilibrium. If so then the sum of the down ward forces, namely F3, F4, F5 and the other downward force by the earth on the bar 236x9.8 N = 2124 N must be equal to sum of upward forces that is F1 and F2 so we have


F1 + F2 = 3900+2990+1960+2124 = 10974 N --------A

Also the sum of the turning moments of all forces about any point let us take about the centre of gravity must be zero
-F1x4.96+3900x3.12 -2990x1.04-1960x3.94+F2x4.96 = 0

[F2 - F1]x4.96= 3109.6+7722.4-12168= -1336

F1-F2 = 269.35 ----------B

2F1 = 11243.35; F1 = 5621.675 N and F2 = 5352.325 N

adding A and B

If the answer is not matching check the arithmetic

I have taken moment about the centre of gravity so the force of earth on the bar which acts at centre of gravity has not been considered.
The formula for taking moments is magnitude of force multiplied by the perpendicular distance from the point on the line of action of the force.
If the force is trying to move the bar anticlockwise the sign is positive and if it moves clockwise it is negative.

Answer 3

Use the law of equlibrium of forces and the law of equilibrium of torgues. These will give you 2 equations, which you can solve with respect to F1 and F2 (2 unknowns).

Remember, these are vector equations. So if the vectors are upward, then they are positive, and if they are downward, they are negative. Also, the torque has a magnitude equal to the product of the force and the arm from the turning point, and is positive if it turns counterclockwise or negative for clocwise. Also, the weight of the beam is W = 236x10 = 2360 N and the length of the beam is D1 + D2 + D3 + D4. The weight acts at the middle of the beam.

The equations should be:

F1 + F2 - F3 - F4 - F5 - W = 0

- F3xD1 - F4x(D1+D2) - F5x(D1+D2+D3) - Wx(D1+D2+D3+D4)/2 + F2x(D1+D2+D3+D4) = 0

When numerical values are substituted, you get:

F1 + F2 - 3900 - 2990 - 1960 = 0
-3900x1.84 - 2990x(1.84+4.16) - 1960x(1.84+4.16+2.90) -
- 2360x(1.84+4.16+2.90+1.02)/2 + F2x(1.84+4.16+2.90+1.02) = 0

F1 + F2 = 8850
9.92xF2 = 54265.6

F1 = 3380 N
F2 = 5470 N

Answer 4

Eq-1. Sum of the forces =0
Eq-2 Sum of the moments = 0 (Moment is a torque or rotational force. It is defined as product between an arm (or radius) times the force. The direction on moment could clockwise or counter clockwise. Just be consistent in signs)

These are the fundamental ideas you need to implement.

Since the forces are all either down or up
use equation 1 (Eq-1)6290N

F1+F2=W+F3+F4+F5
W=mg=mass of the bar times gravitational acceleration

Now lets write a moment equation around point where say F1 is acting

The moment due to F1 will be zero (no radius) and it is convenient.

M(F1)=D1F3 + (D1+D2)F4 + (D1+D2 + D3)F5 +
.5 (D1+D2+D3+D4)W - (D1+D2+D3+D4)F2=0

F2= [D1F3 + (D1+D2)F4 + (D1+D2 + D3)F5]/ (D1+D2+D3+D4)

And we know from above that

F1= W+F3+F4+F5- F2

Just do your math

F2=[(1.84 x3900 +(1.84 + 4.16) 2990 + (1.84 + 4.16+2.9) 1960 +
.5(1.84 + 4.16+2.90+1.02)236 (9.81) ]/(1.84 + 4.16+2.90+1.02)

F2=[(7176)+(17940) +(23245.6) + (22966)]/( 9.92)=
F2=71328/(9.92)= 7190 N

Since we know F2 we can compute F1

F1=236 kg (9.81m/sec^2) + 3900 + 2990 + 1960 - 7190 =
F1=3065 N

(To draw a free body diagram draw a bar supported on each end with arrows pointing upward and labeled F1 and F2. Proportional to scale draw other forces a specified by the problem. They will point downward. Finally draw the force W downward starting at the center of the bar.)

Answer 5

Ok.

First, i'll assume that you can take the weight as being a 2166N force in the middle of the beam.

Your first step is good. Doing free body diagram is always useful.

Now, you have to find enough equations for each of your unkownws. In this case, you can use the following 2:
Sum of forces equals 0
Sum of moments (force times distance) at a given point is 0.

I think you already know how to do the first one. The second one, multiply each force by its distance to one end of the beam and the sum of that will give you 0. Say you do it at the point where F1 is applied, this will give you an equation you can solve right away for F2.

Answer 6

The length of the beam is 9.92 m. Since we are told it is of uniform mass, the center of gravity, C, of the beam will be at 9.92/2 or 4.96 m. This puts F4 1.03 to the right of C, F3 3.13 to the left of C, and F1, of course, is 4.96 to the left of C. Therefore the balance of forces to the left of C will be F3 x 3.13 = F1 x 4.96. Since F3 is 3900 , F1 will be 3900 x 3.13/4.96 = 2461.09N. On the right side of C we have F4 acting at a distance of 1.03 from C and F5 acting at a distance of 1.03 + 2.90 or 3.93. Finally F2 will be acting at a distance of 4.96. Therefore the balance of forces will be given by 2990 x 1.03 + 1960 x 3.93 = F2 x 4.96. Or F2 = 2173.89N.

Answer 7

(considering beam is at rest)
length of the beam = 9.92m
wt acts at middle of the beam i.e. at 4.96m from each end.
All the forces acting are perpendicular to the beam.
Therefore torque = force X distance
Calculate the torque about each end and equate each of them seperately to zero(since beam is at rest).
The eqn of torque about the F1 end will not contain F1.Sly wid F2.
the two eqns will give the values of F1 and F2.

Answer 8

Take a reference point on the beam and balance the following two things:
1. upward force = downward force
2. clockwise moment = anticlockwise moment ( at the reference point)

You have two equations and 2 variables,solve