A heavy door seals a furnace used to heat-treat metal parts. The door\'s weight

Question

A heavy door seals a furnace used to heat-treat metal parts. The door's weight is W=200 lbs which acts through a point G located at the center of the 18" by 18" door. Determine the force supported by cable AB and the reactions at the bearings at points C and D

A heavy door seals a furnace used to heat-treat metal parts. The door's weight is W=200 lbs which acts through a point G located at the center of the 18" by 18" door. Determine the force supported by cable AB and the reactions at the bearings at points C and D

Explanation / Answer

let ED= y"

then from the diagram we have

D= (0,y,0) D'=(0,y-4,0)

C= (0,y+10,0) C'=(0,y+10+4,0)=(0,y+14,0)

B= (0,y+10+24,0)=(0,y+34,0)

the door is in the plane of x=-6

=> A= (-6,y+14,26) G=(-6,y+5,17)

=> AB = (6,y+34-(y+14) , -26) = (6 i + 20 j + -26 k)

DC= (0,y+10-y,0) = 10 j

DA = (-6,y+14-y,26) = (-6 i +14j+26 k)

DG = (-6 ,y+5-y,17) =(-6i+5j+17k)

let the reactions at C be C and that at D be D

then C=(Cx i + Cy j + Cz k ) D=(Dx i + Dy j+ Dz k)

let the tesion in the string be T = T(AB)=T(6 i + 20 j + -26 k)

weight of thr door , W= -200 j

analysing the FBD of the door the forces peresent on the door are T,W,C,D

as it is in equi. T + W + C + D = 0

by equating each component to 0 we get

Tx + Wx + Cx + Dx = 0 => 6T + 0 + Cx +Dx = 0 => 6T+Cx+Dx=0 -------(1)

Ty + Wy + Cy + Dy = 0 => 20T-200+Cy+Dy = 0 => 20T +Cy+Dy=200 ---------(2)

Tz + Wz+Cz+Dz =0 => -26T +0 + Cz +Dz =0 => -26T+ Cz + Dz = 0 -------------(3)

also the door must be in rotational equilibrium

=> moment about any point is zero. Let us take D to be our point.

then moment about D ,Md = DC x C + DG x W + DD x D + DA x T (Since M = r x F )

DC x C = 10 j x (Cx i + Cy j + Cz k ) = 10(Cz i - Cx k)

DG x W = (-6i+5j+17k) x -200 j = 1200 k + 3400 i

DA x T = (-6 i +14j+26 k) x T(6 i + 20 j + -26 k) = T(-1144 i-204 k )

=> (10Cz +3400 - 1144T) i+ (-10Cx +1200 -204 T ) k = 0

=> 10Cz + 3400 -1144 T = 0 ---------- (4) and

-10Cx + 1200 -204T =0------------- (5)

since there are 7 unknowns and only 5 equations to solve this problem is statistically indeterminate.

However in most of the cases we assume bearings do not provide any axial thrust

which implies Cy= Dy = 0

then from equation (2) we have 20T =200 => T=10 lb

equations (1) and (3) transform to Cx + Dx = -60 lb and Cz + Dz = 260 lb

and from equations (4) and (5) we get Cz = 804 lb and Cx = -84 lb

=> from (1) and (3) again we get Dx = 24 lb and Dz = -544 lb

therefore T = 10(6 i + 20 j + -26 k) lb ,magnitude = 333.47 lb

C = (-84 i + 804 k) lb ; magnitude = 808.38 lb

D = (24 i - 544 k) lb ; magnitude = 544.53 lb

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