VECTOR MECHANICS FOR ENGINEERS W/CON >B
VECTOR MECHANICS FOR ENGINEERS W/CON >B
12th Edition
ISBN: 9781260804638
Author: BEER
Publisher: MCG
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Chapter 6.4, Problem 6.159P

The gears A and D are rigidly attached to horizontal shafts that are held by frictionless bearings. Determine (a) the couple M0 that must be applied to shaft DEF to maintain equilibrium, (b) the reactions at G and H.

Chapter 6.4, Problem 6.159P, The gears A and D are rigidly attached to horizontal shafts that are held by frictionless bearings.

Fig. P6.159

(a)

Expert Solution
Check Mark
To determine

The couple M0 that must be applied to shaft DEF to maintain equilibrium.

Answer to Problem 6.159P

The couple M0 that must be applied to shaft DEF to maintain equilibrium is (12.50N.m)i.

Explanation of Solution

Take all vectors along the x axis and y axis as positive.

Radius of gear A is 80mm and radius of gear D is 50mm.

Consider the projection of the gears on yz plane.

The free body diagram of the Gear A and D  is shown below figure1.

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 6.4, Problem 6.159P , additional homework tip  1

Here,MD is the moment of force acting at gear D and J is the tangential force.

Write the expression for the moment at A

MA=F×D (I)

Here, MA is the net moment at A, F is the force, and D is the perpendicular distance between the A and the point where force is experienced.

Above equation implies that net moment at any point is the sum of product of each force acting on the system and perpendicular distance of the force and the point.

The moment at A is due to the external torque of 20N.m and due to the tangential force J.

Thus, write the complete expression of anticlockwise moment MA at A.

MA=20N.mJ(rA) (II)

Here, rA is the radius of gear A.

At equilibrium, the sum of the moment acting at A will be zero.

Write the expression for the total anticlockwise moment acting at A.

MA=20N.mJ(rA)=0 (III)

Write the expression for the moment at D

MD=F×D (IV)

Here, MD is the net moment at D, F is the force, and D is the perpendicular distance between the D and the point where force is experienced.

Above equation implies that net moment at any point is the sum of product of each force acting on the system and perpendicular distance of the force and the point.

The moment at D is due to the couple M0 and tangential force J.

Thus, write the complete expression of anticlockwise moment MD at D.

MD=M0J(rD) (V)

Here, M0 is the magnitude of couple at applied at gear A and rD is the radius of gear D.

At equilibrium, the sum of the moment acting at D will be zero.

Write the expression for the total anticlockwise moment acting at A.

MD=M0J(rD)=0 (VI)

Calculation:

Substitute 80mm for rA in equation (III) to get J.

20N.mJ(80mm)=0

J=20N.m80mm×1m1000mm=250N

Substitute 250N for J and 50mm for rD in equation (VI) to get M0.

(M0)(250N)(50mm×1m1000mm)=0M0=12.50N.m

Since the rotation is in the yz plane , the direction of couple is in x direction.

Therefore, the couple M0 that must be applied to shaft DEF to maintain equilibrium is (12.50N.m)i.

(b)

Expert Solution
Check Mark
To determine

The reaction at G and H.

Answer to Problem 6.159P

The point G experiences a moment of force , MG=(45.5N.m)i and the point H experiences a moment of force, MH=(13.00N.m)i.

Explanation of Solution

Free body diagram of Projection on xz plane Gear A and axle AC is shown in figure 2.

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 6.4, Problem 6.159P , additional homework tip  2

Here, is the tangential force acting on the gear, B is the force acting at point B , C is the force acting at point C.

From figure 2, write the equation of net moment about B.

MB=F×D (VII)

Here, MB is the net moment at B, F is the force, and D is the perpendicular distance between the B and the point where force is experienced.

Above equation implies that net moment at any point is the sum of product of each force acting on the system and perpendicular distance of the force and the point.

The moment at B is due to the tangential 250N.m on the gear and the force C.

Thus, write the complete expression of anticlockwise moment MB at B.

MB=C(dBC)+(250N)(dJB) (VIII)

Here, dBC is the perpendicular distance between the points B and C and dJB is the perpendicular distance between the force J and point B.

At equilibrium, the sum of the moment acting at B will be zero.

Write the expression for the total anticlockwise moment acting at A.

MB=C(dBC)+(250N)(dJB)=0 (IX)

Write the expression for the moment at D

MC=F×D (X)

Here, MC is the net moment at C, F is the force, and D is the perpendicular distance between the C  and the point where force is experienced.

Above equation implies that net moment at any point is the sum of product of each force acting on the system and perpendicular distance of the force and the point.

At equilibrium, the sum of the moment acting at C will be zero.

Write the expression for the total anticlockwise moment acting at A.

MC=B(dBC)+(250N)(dJC)=0 (XI)

Here, dBC is the perpendicular distance between the points B and C and dJC is the perpendicular distance between the force J and point C.

The free body diagram of the projection on xz plane with Gear D and axle DF is shown in figure3.

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 6.4, Problem 6.159P , additional homework tip  3

Here, E is the force acting at point E and F is the force acting at point F.

From figure 3, write the equation of net moment about E.

ME=F×D (XII)

Here, ME is the net moment at E, F is the force, and D is the perpendicular distance between the E and the point where force is experienced.

Above equation implies that net moment at any point is the sum of product of each force acting on the system and perpendicular distance of the force and the point.

The moment at E is due to the tangential 250N.m on the gear and the force F.

Thus, write the complete expression of anticlockwise moment ME at E.

ME=F(dFE)+(250N)(dJF) (XIII)

Here, dFE is the perpendicular distance between the points F and E and dJF is the perpendicular distance between the force J and point F.

At equilibrium, the sum of the moment acting at E will be zero.

Write the expression for the total anticlockwise moment acting at E.

ME=+F(dFE)(250N)(dJF)=0 (XIV)

Write the expression for the moment at F

MF=F×D (XV)

Here, MF is the net moment at C, F is the force, and D is the perpendicular distance between the F  and the point where force is experienced.

Above equation implies that net moment at any point is the sum of product of each force acting on the system and perpendicular distance of the force and the point.

At equilibrium, the sum of the moment acting at F will be zero.

Write the expression for the total anticlockwise moment acting at A.

MF=E(dEF)(250N)(dJF)=0 (XVI)

Here, dBC is the perpendicular distance between the points B and C and dJC is the perpendicular distance between the force J and point C.

Consider the projection at yz plane.

The free body diagram of the Bracket BG.

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 6.4, Problem 6.159P , additional homework tip  4

Here, B is the force at point B, E is the force at point E , MG is the moment at point G and G is the force at point G.

Write the expression for the total force along z direction.

Fz=GE+B (XVII)

Since in this direction net force is equal to zero. Equate above equation to zero.

Fz=GE+B=0 (XVIII)

Since total moment of force about x axis is zero, moment of force about E is zero.

Write the equilibrium moment of force about E.

MG+B(130mm)=0 (XIX)

The free body diagram of the Bracket CH.

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 6.4, Problem 6.159P , additional homework tip  5

Here, C is the force at point C, F is the force at point F , MH is the moment at point H and H is the force at point H.

Write the expression for the total force along z direction.

Fz=HF+C (XX)

Since in this direction net force is equal to zero. Equate above equation to zero.

Fz=HF+C=0 (XXI)

Since total moment of force about x axis is zero, moment of force about E is zero.

Write the equilibrium moment of force about E.

MH+C(130mm)=0 (XXII)

Calculation:

Substitute 100mm for dBC and 40mm for dJB in equation (VIII) to get C.

C(100mm)+(250N)(40mm)=0

C=(250N)(40mm)(100mm)=100N

Substitute 140mm for dJC and 100mm for dBC in equation (XI) to get B.

B(100mm)+(250N)(140mm)=0

B=(250N)(140mm)(100mm)=350N

Substitute 100mm for dFE and 40mm for dJE in equation (XIV) to get E.

F(100mm)(250N)(40mm)=0

F=(250N)(40mm)100mm=100N

Substitute 140mm for dJF and 100mm for dFE in equation (XI) to get E.

E(100mm)(250N)(140mm)=0

E=(250N)(140mm)100mm=350N

Substitute 350N for E and B in equation (XVIII)to get G.

G350N+350N=0G=0N

Substitute 350N for B in equation (XIX) to get MG.

MG=(350N)(130mm×1m1000m)=45.5Nm

The negative sign indicate that it is directed along x direction.

Substitute 100N for F and C to get H.

H100N+100N=0H=0N

Substitute 100N for C in equation (XXII) to get MH.

MH=(100N)(130mm×1m1000m)=13.00Nm

The positive value indicate that it is directed along x direction.

Substitute 100N for F and C to get H.

H100N+100N=0H=0N

Therefore, the net force at G and H is zero. The moment of force about GMG=(45.5Nm)i and moment of force about H , MH=(13.00N.m)i

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Chapter 6 Solutions

VECTOR MECHANICS FOR ENGINEERS W/CON >B

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