Engineering Mechanics Archives -

Question: Determine the horizontal and vertical components of reaction at the pin A and the reaction at the roller support B required for equilibrium of the truss.
[A].

Ax = 0, Ay = 333 lb, NB = 267 lb

[B].

Ax = 462 lb, Ay = 66.7 lb, NB = 533 lb

[C].

Ax = 267 lb, Ay = 223 lb, NB = 377 lb

[D].

Ax = 154.0 lb, Ay = 333 lb, NB = 308 lb

Answer: Option D

Explanation:

No answer description available for this question.

Determine the horizontal and vertical components of reaction at the pin A and the reaction at the roller support B required for equilibrium of the truss. Read More »

Engineering Mechanics, Equilibrium Of A Rigid Body

Compute the horizontal and vertical components of force at pin B. The belt is subjected to a tension of T=100 N and passes over each of the three pulleys.

Engineering Mechanics, Equilibrium Of A Rigid Body / admin

Question: Compute the horizontal and vertical components of force at pin B. The belt is subjected to a tension of T=100 N and passes over each of the three pulleys.
[A].

Bx = 0 N, By = 141.4 N

[B].

Bx = -15.89 N, By = 120.7 N

[C].

Bx = 20.7 N, By = 157.3 N

[D].

Bx = 0, By = 100.0 N

Answer: Option B

Explanation:

No answer description available for this question.

Compute the horizontal and vertical components of force at pin B. The belt is subjected to a tension of T=100 N and passes over each of the three pulleys. Read More »

Engineering Mechanics, Equilibrium Of A Rigid Body

According to parallel axis theorem, the moment of inertia of a section about an axis parallel to the axis through centre of gravity (i.e. IP) is given by(where, A = Area of the section, IG = Moment of inertia of the section about an axis passing through its C.G., and h = Distance between C.G. and the parallel axis.)

Engineering Mechanics, Mechanical Engineering / admin

Question: According to parallel axis theorem, the moment of inertia of a section about an axis parallel to the axis through centre of gravity (i.e. IP) is given by(where, A = Area of the section, IG = Moment of inertia of the section about an axis passing through its C.G., and h = Distance between C.G. and the parallel axis.)
[A].

IP = IG + Ah2

[B].

IP = IG – Ah2

[C].

IP = IG / Ah2

[D].

IP = Ah2 / IG

Answer: Option A

Explanation:

No answer description available for this question.

According to parallel axis theorem, the moment of inertia of a section about an axis parallel to the axis through centre of gravity (i.e. IP) is given by(where, A = Area of the section, IG = Moment of inertia of the section about an axis passing through its C.G., and h = Distance between C.G. and the parallel axis.) Read More »

Engineering Mechanics, Mechanical Engineering

Moment of inertia of a triangular section of base (b) and height (h) about an axis passing through its vertex and parallel to the base, is __________ than that passing through its C.G. and parallel to the base.

Engineering Mechanics, Mechanical Engineering / admin

Question: Moment of inertia of a triangular section of base (b) and height (h) about an axis passing through its vertex and parallel to the base, is __________ than that passing through its C.G. and parallel to the base.
[A].

nine times

[B].

six times

[C].

four times

[D].

two times

Answer: Option A

Explanation:

No answer description available for this question.

Moment of inertia of a triangular section of base (b) and height (h) about an axis passing through its vertex and parallel to the base, is __________ than that passing through its C.G. and parallel to the base. Read More »

Engineering Mechanics, Mechanical Engineering

Moment of inertia of a triangular section of base (b) and height (h) about an axis through its base, is

Engineering Mechanics, Mechanical Engineering / admin

Question: Moment of inertia of a triangular section of base (b) and height (h) about an axis through its base, is
[A].

bh3/4

[B].

bh3/8

[C].

bh3/ 12

[D].

bh3/ 36

Answer: Option C

Explanation:

No answer description available for this question.

Moment of inertia of a triangular section of base (b) and height (h) about an axis through its base, is Read More »

Engineering Mechanics, Mechanical Engineering

Moment of inertia of a triangular section of base (b) and height (h) about an axis passing through its C.G. and parallel to the base, is

Engineering Mechanics, Mechanical Engineering / admin

Question: Moment of inertia of a triangular section of base (b) and height (h) about an axis passing through its C.G. and parallel to the base, is
[A].

bh3/4

[B].

bh3/8

[C].

bh3/12

[D].

bh3/36

Answer: Option D

Explanation:

No answer description available for this question.

Moment of inertia of a triangular section of base (b) and height (h) about an axis passing through its C.G. and parallel to the base, is Read More »

Engineering Mechanics, Mechanical Engineering

Moment of inertia of a hollow circular section, as shown in the below figure about an axis perpendicular to the section, is __________ than that about X-X axis.

Engineering Mechanics, Mechanical Engineering / admin

Question: Moment of inertia of a hollow circular section, as shown in the below figure about an axis perpendicular to the section, is __________ than that about X-X axis.
[A].

two times

[B].

same

[C].

half

Answer: Option A

Explanation:

No answer description available for this question.

Moment of inertia of a hollow circular section, as shown in the below figure about an axis perpendicular to the section, is __________ than that about X-X axis. Read More »

Engineering Mechanics, Mechanical Engineering