Problem 4 A 50 kg uniform horizontal beam of length 3.0 m is attached to a brick

Question

Problem 4 A 50 kg uniform horizontal beam of length 3.0 m is attached to a brick wall. A cable at the far end makes an angle of 60 degrees with the horizontal a) What is the weight of the beam? b) The beam is uniform, which means that it has the same density throughout not like a baseball bat that has a funny shape and more mass on one end than on the other. With that in mind, where would you expect the center of gravity of the beam to be? c) What is the equation for the y component of the tension in the cable? (what I d) Using what you know about torque, calculate what the y component of the e) Using what you know about net force, calculate what the y component of the mean is Ty = T sin() or T,-T cos(9) which one is it?) tension is force from the wall on the beam is. 60

Explanation / Answer

a)

weight W = mg = 50*9.8 = 490 N

(b)

the center of mass is at the half way L/2 ( i.e 1.5 m )

(c)

T is making an theta = 60 with horizontal

vertical component of tension Ty = T*sintheta

(d)

net torque about the wall = 0

Ty*L - Mg*L/2 = 0

Ty = Mg/2

Ty = 490/2 = 245 N

(e)

along verticla Fnet = 0

Ty - Mg + Fy = 0

Fy = Mg - Ty

Fy = Mg - Mg/2

Fy = Mg/2 = 245 N <<<<------ANSWER

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