Consider the flagpole in the figure below. If the flagpole has a mass of 21 kg a

Question

Consider the flagpole in the figure below. If the flagpole has a mass of 21 kg and length 10 m and the angle the cable makes with the pole is theta = 20, what are the magnitude and direction of the force exerted by the hinge (at point P) on the flagpole? Assume the mass of the pole is distributed uniformly.

Consider the flagpole in the figure below. If the flagpole has a mass of 21 kg and length 10 m and the angle the cable makes with the pole is theta = 20½, what are the magnitude and direction of the force exerted by the hinge (at point P) on the flagpole? Assume the mass of the pole is distributed uniformly. Magnitude direction counterclockwise from the + x ? axis

Explanation / Answer

The net force must be zero in the x and y direction

sum of the forces in the x direction=0= Hx+ Tcos(theta)

sum of the forces in the y direction = 0 = Hy-mg+Tsin(theta)

Substitute the know values, theta is 20 degrees, mass of the flag pole is 21 kg, and gravity is 9.81 m/s^2

0= Hx+ Tcos(20)

0 = Hy-(21 kg)9.81 m/s^2+Tsin(theta)

The sum of the moments of the system needs to be included since, we have 2 equations and 3 unknowns,

sum of the moment at H=0= Tsin(theta)(L) - mg(L/2)

Substitute the know values, theta is 20 degrees, mass of the flag pole is 21 kg, length, L is 10 m, and gravity is 9.81 m/s^2

0= T sin(20)(10 m) - (21 kg )(9.81 m/s^2 )(10 m /2)

T=(21 kg )(9.81 m/s^2 )(10 m /2)/(sin(20)(10 m))=301.2N

Substitute back into the equations

Hy=(21 kg)9.81 m/s^2-(301.2 N)sin(20)=103 N

Hx= -(301.2 N)cos(20)=- 283 N

your total magnitude would be sqrt(Hx^2+Hy^2)=sqrt(283^2+103^2)?301.2 N is your magnitude

your direction would be tan(theta1)=Hy/Hx

tan^-1(103/(-283))=-20degrees would be your direction.

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