Problem 10.62: The farmyard gate. A gate 4.00 m wide and 2.00 m high weighs 480N

Question

Problem 10.62: The farmyard gate.

A gate 4.00 m wide and 2.00 m high weighs 480N . Its center of gravity is at its center, and it is hinged at A andB. To relieve the strain on the top hinge, a wire CD is connected as shown in the figure (Figure 1) . The tension in CD is increased until the horizontal force at hinge A is zero.

Part A

What is the tension in the wireCD?

Part B

What is the magnitude of the horizontal component of the force at hinge B?

Part C

What is the combined vertical force exerted by hinges A andB?

Problem 10.62: The farmyard gate. A gate 4.00 m wide and 2.00 m high weighs 480N . Its center of gravity is at its center, and it is hinged at A andB. To relieve the strain on the top hinge, a wire CD is connected as shown in the figure (Figure 1) . The tension in CD is increased until the horizontal force at hinge A is zero. no title provided Part A What is the tension in the wireCD? Part B What is the magnitude of the horizontal component of the force at hinge B? Part C What is the combined vertical force exerted by hinges A andB?

Explanation / Answer

Width, w = 4 m

Height, h = 2 m

Weight, W = 480 N

Angle, ? = 30 deg

Solution:

(a)

?? = 0 about 'B':

( T sin 30 ) * 4 + ( T cos 30 * 2 ) - W * ( 4 / 2 ) = 0

T = 480 * 2 / ( 4 * 0.5 + 2 * 0.866 )

   = 257.23 N

Ans:

Tension, T = 257.23 N

(b)

?Fx = m ax

Hbh - T cos 30 = 0

Hbh = 257.3 * cos 30

        = 222.77 N

Ans:

Horizontal component of force at B:

Hbh = 222.77 N

(c)

?Fy = m ay

Hav + Hbv + T sin 30 - W = 0

Hav + Hbv = W - T sin 30

                   = 480 - ( 257.23 N * sin 30 )

                   = 351.385 N

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