For part (a) I am getting two different T values. I got 846 N and 3360 N using t

Question

For part (a) I am getting two different T values. I got 846 N and 3360 N using two different calculation method. I am not sure which method I did was correct if any.

Thanks,

A 1175 N uniform boom at phi = 69.0 degree to the vertical is supported by a cable at an angle theta = 21.0 degree to the horizontal as shown in Figure 3. The boom is pivoted at the bottom, and an object of weight m = 2150 N hangs from its top. Find (a) the tension in the support cable and (b) the components of the reaction force exerted by the floor on the boom.

Explanation / Answer

Given that

phi = 69.0 degree

m = 2150 N

Thita = 21.0

(A)
Take moments (force multiplies by perpendicular distance) about the bottom hinge

m*g(l*cos(phi)) + W(l/2*cos(phi)) - T(3*l/4) = 0

T = 4/3[m*g*cos(phi) + (W/2)*cos(phi)]

= 4/3[2150 N*cos(69.0) + (1175 N/2)*cos(69.0)]

T = 1308.01 N

(b)
The equation of static equilibrium for boom is

SigmaFx = 0

Rx - T*cos(21.0) = 0

Rx = 1308.01*cos(21.0)

Rx = 1221.13

This reaction force is directed towards right
Also

Ry + T*sin(21.0) - T' - 1175 = 0

Ry = W + mg - tsin(thita)

= 1175 + 2150 - 1308.01*sin21.0

= 2856 N

This reaction force is upward.

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