A shop sign weighs 201 N and is supported by a uniform 129 N beam of length =1.70 m. A guy wire is connected D=1.36 m from the backboard. The angle between the wall and the board = 39.2 degrees.
What is the tension in the guy wire?
What is the horizontal and vertical force exerted by the hinge on the beam?
I have drawn an FBD and used these three equations to solve this problem, but I keep getting the wrong answer.
W(center of mass of beam) times r(center of mass)= Tension (sin theta) times r (tension)
T (y direction) + force on wall = W (center of mass)
T (x direction) = T (on wall)
Are these the right equations? Am I doing this right? What should I plug in for the variables?
2007-01-16 07:50:47 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Physics
I saw you before - I told you something was wrong with this. Nothing weighs a force!
The problem is unsolvable the way it is stated.
2007-01-16 08:21:40 · answer #1 · answered by Dr Dave P 7 · 0⤊ 0⤋
I am not clear on where exactly the guy wire is located. Can you try to be more specific? Once you specify this, there are 3 equations, as you mention:
Sum of F in x direction = 0
Sum of F in y direction = 0
Sum of torque = 0.
I am going to assume that the wire is attached along the beam, 1.36m away from the wall.
My equations therefore become:
Tx = Hx (T = Tension, H = hinge)
or T cos 39.2 = Hx
Ty + Hy = 129 + 201 (assume for the moment Hy is upwards).
or T sin 39.2 + Hy = 330
Ty(1.36) = (201)(1.70) + (129)(0.85)
or Ty = 331.875 N.
Working backwards, we quickly find Hy = 1.875 N (downwards), T = 525 N, Tx = 406.9 N, and Hx = 406.9 N to the right.
This should be a good start. I'd have to know more about the guy wire to be more specific.
2007-01-16 08:25:56 · answer #2 · answered by Blueearth423 2 · 0⤊ 0⤋
by way of fact the shell is max top, it has an preliminary velocity of 0. Then to drop sixteen.40 8.... m, x = a million/2 gt^2, the place x is sixteen.40 8 and g is 9.8 m/sec2. this could be a pair of million.4 sec. the cost whilst it hits the floor is v= gt, the place g is 9.8 m/sec^2 and t = a million.4 sec.
2016-12-16 06:11:33 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Where's the center of mass of the sign?
2007-01-16 08:21:49 · answer #4 · answered by injanier 7 · 0⤊ 0⤋
Is there a way you can scan an image of the problem? No one can tell what is going on from the problem statement.
2007-01-16 08:19:06 · answer #5 · answered by daedgewood 4 · 0⤊ 0⤋