Physics question:
The black line (B) represents a stand
The Red line (A) represents a beam
The blue line (C) represents a very sturdy strong structure attached to the earth.
x and y are the distances from each end of the red beam to the point where it touches the black stand.
The question is: how much force is being exerted down on the top of the black stand?
>Force
>No mass
Wut
the red beam has a mass of Mass(A)
It's zero it's not moving, no acceleration
Acceleration comes from gravity
What is normal force
Acceleration due to gravity * (Mass(A)+Mass(B))
C is attached to earth so it's upside down. They specified earth so a=9.8m/s and since nothing is on "top" of B from the inverted perspective, the force is (mass B)*a
>The blue line (C) represents a very sturdy strong structure attached to the earth.
Is the pic upside down then?
>on the top of the black stand?
So at the very bottom of the black line in the pic?
To clarify:
C is a very strong object ABOVE A and B that is attached (off screen) to the earth, like the top opening of a cave, think of it as an immovable object.
The picture is not upside down.
If C weren't present, A would topple off B.
B is pushing on C, C is in B's way and stopping it falling off.
Gravity is indeed affecting all the objects and object A and B are pushing down on object C but, the normal force, like said, is pushing back on the force from gravity, canceling it out, making acceleration 0. Since 0 multiplied by anything is 0, and F=ma, the force is gonna be 0.
Mass(A) * (1 + (x-y)/(x+y) ) * g
It's lacking information, OP. Where's the earth? how are the bars connected? (do they allow rotation?)
Without any numbers you can't quantify the force dipshit
variables dipshit
>B is pushing on C, C is in B's way and stopping it falling off.
Explain this. How is B pushing on C?
Fucking make a sum of momentums @ the point of contact between the blue and the red lines. It will give out that mass(a)*g*((x+y)/2)=F*y.
Solve for F
Normal force pushes back on the gravitational force that C is putting on B
sorry, its supposed to be A is pushing on C
because A is supported outside it's centre of gravity, gravity is applying a torque to A.
if C weren't present, A would rotate anti clockwise and fall off B
not sure I follow
they all have momentum of 0 because nothing is moving.
Ok but nothing is holding up B right?
Is the problem assuming t=0 just before B starts falling because of gravity?
Since you clarify, it's solvable. Think of a linear load being applied on A which corresponds to its weight. Lets call whis 'w', which is measured in "force/distance", like N/m, for instance.
Now, just consider you have w*x as the weight on the left side of B, applied on a distance of y+x/2 from the left. Same way, you have a w*y load applied on y/2 from the left.
Now, archimedes principle says that F1*d1 = F2*d2, F1 and F2 being forces and d1 and d2 distances from the same point.
That given, alto consider you have a reaction force of R, applied in a 'y' distance from the right.
now, you have the following equation, from archimedes principle:
(w*x)*(y+x/2) + (w*y)*y/2 + R*y = 0
that gives R = -(w*(x*(y+x/2) + y*y/2))/y
F=ma just means those quantities must balance.
there is a force upwards that counters the force downwards therefore no acceleration.
F(up)+F(down)=0, so F(up)=-F(down) bearing in mind that these are vector quantities.
The point where B meets a is a fulcrum. The mass of the portion of A that lies to the left of the fulcrum is the length of A times x/(x+y), and the length of the portion to the right is y(x+y).
Now you have to solve using the formula for torque this because it is swinging around a fulcrum. The torque need to balance like Newtons second law. Be sure to work out the moment of inertia using the lengths I gave you.
I won't do your homework for you user, but I've given you a real liberal head start.