A 540-N uniform rectanqular sign 4.00 m wide and 3.00 m high is suspended from a
ID: 1366408 • Letter: A
Question
A 540-N uniform rectanqular sign 4.00 m wide and 3.00 m high is suspended from a horizontal, 6.00-m-long, uniform, 120-N rod as indicated in the figure below. The left end of the rod s supported by a hinge and the right end is supported by a thin cable making a 30.0° angle with the vertical. (Assume the cable is connected to the very end of the 6.00-m-long rod and that there are 2.00 m separating the wall from the sign.) 0 ICE CREAM SHOP (a) Find the (magnitude of the) tension T in the cable (b) Find the horizontal and vertical components of the force exerted on the left end of the rod by the hinge. (Take up and to the right to be the positive directions. Indicate the direction with the sign of your answer.) horizontal component IN ertical component nt Need Help? Read ItMaster ItExplanation / Answer
m1 = 10000 kg
m2 = 20000 kg
u = 0.425
minimum acceleration = -u*g
from equation of motion
v^2 - u^2 = 2*a*x
u = initial velocity = 10 m/s
v = final velocity = 0
a = u*m1*g
v^2 - u^2 = 2*u*g*x
0^2 - 10^2 = -0.425*9.8*x
x = 24 m <<<<<<<<--------------answer
W1 = 540 N
r1 = 2+2 = 4 m fromhinge
w2 = 120 N
r2 = L/2 = 3 m from hinge
In equilibrium net torque = 0
T*cos30*L = W1*r1 + w2*r2
T*cos30*6 = 540*4+120*3
T = 484.97 N <<---------------answer
+++++++++++++
net force = 0
along vertical
Fy - w1 - w2 + T*cos30 = 0
Fy - 540 - 120 + (484.97*cos30) = 0
Fy = 240 N <<<<<-----------------------answer
along horizantal
Fx - T*sin30 = 0
Fx = 242.5 N <<<<<<<<<<<-----------------------answer
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