Hard Physics Question!! I am totally lost....?

Question

A shop sign weighing 201 N is supported by a uniform 129 N beam of length L = 1.70 m. The wire is connected D = 1.36 m from the backboard.
Find the tension in the wire. Assume the angle= 39.2°
Find the horizontal and vertical forces exerted by the hinge on the beam. (Use up as the positive direction.)

Where do you start on this?? I am so frustrated. My teacher gives us hardly any help on questions. He speaks abstractly and is confusing when does.

Answer 1

First, sum up the forces on the beam-sign system. They should equal zero f = ma = sum(all forces) = 0. Why? Because the system is not accelerating, it is fixed to the wall (backboard) and to the wire at the other end.

Tension (T) is just one force, but this acts along the wire; so the vertical component of tension = Tv = T sin(39.2 deg). The corresponding horizontal tension component = Th = T cos(39.2 deg). So there are two forces we need to know...the vertical and horzontal components of the tension (T).

What other forces are there? Well, there's weight...of the system (discounting the wire).... So, we need to break W = Mg into its vertical and horizontal parts Wh = 0 and Wv = W = Mg because there is no horizontal component for weight. M = the masses of the sign and beam, and g = 9.81 m/sec^2 on Earth's surface.

Note that weight acts on a center of mass (M), which for your problem would be half way out on the beam for the vertical component of weight (Wv). Thus, as Wv X D/2 = wv X D, from the moment arms on the beam, we have wv = Wv/2 on the beam where the wire is attached and also where the beam is attached to the backboard.

There is yet one more force, the force of the beam against the backboard. This exists because the beam is stationary and therefore the net forces on it sum to zero. The only horizontal force we've found thus far is Th, which is transmitted along the beam into the wall. Thus, Bh = Th + Wh = Th because Wh = 0; where Bh is the backboard force counteracting the tension horizontal force. The counteracting vertical force at the wall is Bv = Wv/2 = wv.

Now, we've got all the forces broken out into their vertical and horizontal components. Next we sum them up where the wire is attached to the system so they equal zero. From the vertical forces we have Wv/2 = Tv = T sin(39.2); so that 2T = Wv/sin(39.2) = 330/sin(39.2). You can do the math.

Note the beam length (L) has no bearing on this problem. But we do invoke that the total weight (W = Wv) is distributed equally over the two ends of the beam (where it's attached to the backboard and where it's attached to the wire). We got this from knowing that a force X distance (i.e., moment) can be balanced according to Fd = fD so that F(d/D) = f which means that f at distance D is equivalent to F at distance d. In your problem F = Wv(d/D) = Wv(d/2d) = wv = f = Wv/2 as noted earlier.

PS: Contrary to what one answerer wrote, weight is a force...the force of gravity as determined in classical physics by F = ma = mg = W; where all we do is substitute a with g the acceleration due to gravity to convert force (F) to weight (W).

Answer 2

I'm not exactly sure what this shop sign looks like from your description.

How exactly is this sign attached to this beam?

You would start this problem by drawing a vector diagram. Sum the forces in the vertical and horizontal directions and then use trigonometry to find the force along the wire.

To find the forces excerted by the hinge on the beam, remember, they must be such that static equilibrium is achieved. For example. If I push on a wall with a force of 500N, and the wall doesn't go anywhere, then we are in static equilibrium and the nt force of the me wall system is 0. So F_wall + F_me=0

F_wall + 500N = 0
F_wall = -500N

Answer 3

1).Find the co-ordinates of the point A on the line x= -3 such that the line joining A to B(5,3) is perpendicular to the line 2x+5y =12.

2). Using the vertices A(2a,2b), B(-2c, 0), C(2c,0).
(i) prove that the medians of a triangle are concurrent. (a median is a line joining a vertex to a midpoint of the opposite sides; the lines are concurrent if they have a common point of intersection);
(ii) show that the centroid is 2a/3, 2b/3. (a centroid is the point of intersection of the medians).

Answer 4

i don't understand they are each in a matching way confusing to me. Physics probable better desirable although, yet you probable have themes with making formula out of formula then this is larger to be looking out chemistry, simply by fact physics have lots of formula

Answer 5

the answer is 2

Answer 6

Something can't WEIGH newtons, they are a measure of force, not weight!