A 520-N uniform rectangular sign 4.00 m wide and 3.00 m high is suspended from a

Question

A 520-N uniform rectangular 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 is 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.)

(a) Find the (magnitude of the) tension T in the cable.
________ N

(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.)

4.–/1 pointsSerCP10 8.P.030.

One end of a uniform = 3.20-m-long rod of weight w is supported by a cable at an angle of

= 37°

with the rod. The other end rests against a wall, where it is held by friction (see figure). The coefficient of static friction between the wall and the rod is

s = 0.530.

Determine the minimum distance x from point A at which an additional weight w (the same as the weight of the rod) can be hung without causing the rod to slip at point A.
_______ m

Solve both problems please

Explanation / Answer

<< Find the tension, T, in the cable.>>

Take summation of moments about the hinge. Since the system is in equilibrium, then

M = 0

520(4) + 120(3) - T(cos 30)(6) = 0

where T = tension in the cable

Solving for "T"

T(cos 30)(6) =520(4) + 120(3)

T = 469.57 N

<< Find the horizontal and vertical components of force exerted on the left end of the rod by the hinge. (Take up and to the right to be the positive directions.) >>

Summation of vertical forces = 0, i.e.,

V = 0

V +469.57 (cos 30) - 520 - 120 = 0

Solving for "V"

V = 233.34 N

Summation of horizontal forces = 0, i.e.,

H = 0

H - 469.57(sin 30) = 0

H = 234.78 N

Let T be the tension in the cable and N the normal force at the wall, with friction given by FR µsN. Then:

Fx : N T cos() = 0

Fy : T sin() + FR 2 × w = 0, and

: T × Lsin() w × L 2 w × x = 0,

where torque has been calculated around point A. From the first two equations we have:

2 × w = T sin() + FR T sin() + µs × T cos() = T sin() 2 × w /1 + µs × cot() ,

while from the third equation we find

T sin() = w × ( 1/2 + x/L ) .

Putting these together we have

x L ( 2/1 + µs × cot() 1/2 )

= 3.20 * (2/1+0.530*1.327) - 1/2

8.15 m

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