Engineering Mechanics

Engineering Mechanics | Mechanical Engineering

The angle of the inclined plane at which a body just begins to slide down the plane, is called helix angle.

Question: The angle of the inclined plane at which a body just begins to slide down the plane, is called helix angle.
[A].
True
[B].
False

Answer: Option B

Explanation:

No answer description available for this question.

Read More The angle of the inclined plane at which a body just begins to slide down the plane, is called helix angle.
Engineering Mechanics | Mechanical Engineering

A force acting in the opposite direction to the motion of the body is called force of friction.

Question: A force acting in the opposite direction to the motion of the body is called force of friction.
[A].
Agree
[B].
Disagree

Answer: Option A

Explanation:

No answer description available for this question.

Read More A force acting in the opposite direction to the motion of the body is called force of friction.
Engineering Mechanics | Equilibrium Of A Rigid Body

A Russell’s traction is used for immobilizing femoral fractures C. If the lower leg has a weight of 8 lb, determine the weight W that must be suspended at D in order for the leg to be held in the position shown. Also, what is the tension force F in the femur and the distance which locates the center of gravity G of the lower leg? Neglect the size of the pulley at B.

Question: A Russell’s traction is used for immobilizing femoral fractures C. If the lower leg has a weight of 8 lb, determine the weight W that must be suspended at D in order for the leg to be held in the position shown. Also, what is the tension force F in the femur and the distance which locates the center of gravity G of the lower leg? Neglect the size of the pulley at B.
[A].
x = 1.44 ft, w = 10.8 lb, F = 12.61 lb
[B].
x = 1.33 ft, w = 15.76 lb, F = 20.0 lb
[C].
x = 1.56 ft, w = 9.75 lb, F = 12.69 lb
[D].
x = 0.869 ft, w = 6.44 lb, F = 5.03 lb

Answer: Option A

Explanation:

No answer description available for this question.

Read More A Russell’s traction is used for immobilizing femoral fractures C. If the lower leg has a weight of 8 lb, determine the weight W that must be suspended at D in order for the leg to be held in the position shown. Also, what is the tension force F in the femur and the distance which locates the center of gravity G of the lower leg? Neglect the size of the pulley at B.
Engineering Mechanics | Equilibrium Of A Particle

Determine the magnitudes ofthe forces P, R, and F required for equillibrium of point O.

Question: Determine the magnitudes ofthe forces P, R, and F required for equillibrium of point O.
[A].
R = 238 N, F = 181.0 N, P = 395 N
[B].
R = 1340 N, F = 740 N, P = 538 N
[C].
R = 419 N, F = 181.0 N, P = 395 N
[D].
R = 409 N, F = 504 N, P = 1099 N

Answer: Option D

Explanation:

No answer description available for this question.

Read More Determine the magnitudes ofthe forces P, R, and F required for equillibrium of point O.
Engineering Mechanics | Equilibrium Of A Particle

Determine the tension developed in cables OD and OB and the strut OC, required to support the 500-lb crate. The spring OA has an unstretched length of 0.2 ft and a stiffness of kOA = 350lb/ft. The force in the strut acts along the axis of the strut.

Question: Determine the tension developed in cables OD and OB and the strut OC, required to support the 500-lb crate. The spring OA has an unstretched length of 0.2 ft and a stiffness of kOA = 350lb/ft. The force in the strut acts along the axis of the strut.
[A].
Fob = 289 lb, Foc = 175.0 lb, Fod = 131.3 lb
[B].
Fob = 86.2 lb, Foc = 175.0 lb, Fod = 506 lb
[C].
Fob = 375 lb, Foc = 0, Fod = 375 lb
[D].
Fob = 664 lb, Foc = 175.0 lb, Fod = 244 lb

Answer: Option B

Explanation:

No answer description available for this question.

Read More Determine the tension developed in cables OD and OB and the strut OC, required to support the 500-lb crate. The spring OA has an unstretched length of 0.2 ft and a stiffness of kOA = 350lb/ft. The force in the strut acts along the axis of the strut.
Engineering Mechanics | Equilibrium Of A Particle

Determine the force in each strut and tell whether it is in tension or compression.

Question: Determine the force in each strut and tell whether it is in tension or compression.
[A].
Fab = 1.76 lb T, Fac = 5.00 lb T, Fad = 3.53 lb C
[B].
Fab = 11.47 lb T, Fac = 25.0 lb C, Fad = 14.97 lb C
[C].
Fab = 11.47 lb C, Fac = 25.0 lb T, Fad = 14.97 lb C
[D].
Fab = 1.76 lb C, Fac = 5.00 lb T, Fad = 3.53 lb C

Answer: Option C

Explanation:

No answer description available for this question.

Read More Determine the force in each strut and tell whether it is in tension or compression.
Engineering Mechanics | Equilibrium Of A Particle

A continuous of total length 4 m is wrapped around the small frictionless pulleys at A, B, C, and D. If the stiffness of each spring is k = 500 N/m and each spring is stretched 300 mm, determine the mass m of each block. Neglect the weight of the pulleys and cords. The springs are unstretched when d = 2 m.

Question: A continuous of total length 4 m is wrapped around the small frictionless pulleys at A, B, C, and D. If the stiffness of each spring is k = 500 N/m and each spring is stretched 300 mm, determine the mass m of each block. Neglect the weight of the pulleys and cords. The springs are unstretched when d = 2 m.
[A].
m = 153.0 kg
[B].
m = 15.60 kg
[C].
m = 4.75 kg
[D].
m = 30.5 kg

Answer: Option B

Explanation:

No answer description available for this question.

Read More A continuous of total length 4 m is wrapped around the small frictionless pulleys at A, B, C, and D. If the stiffness of each spring is k = 500 N/m and each spring is stretched 300 mm, determine the mass m of each block. Neglect the weight of the pulleys and cords. The springs are unstretched when d = 2 m.
Engineering Mechanics | Equilibrium Of A Particle

Determine the force F needed to hold the 4-kg lamp in the position shown.

Question: Determine the force F needed to hold the 4-kg lamp in the position shown.
[A].
F = 39.2 N
[B].
F = 68.0 N
[C].
F = 34.0 N
[D].
F = 19..62 N

Answer: Option A

Explanation:

No answer description available for this question.

Read More Determine the force F needed to hold the 4-kg lamp in the position shown.
Engineering Mechanics | Equilibrium Of A Particle

A “scale” is constructed with a 4-ft-long cord and the 10-lb block D. The cord is fixed to a pin at A and passes over two small pulleys at B and C. Determine the weight of the suspended block E if the system is in equilibrium when s = 1.5 ft.

Question: A “scale” is constructed with a 4-ft-long cord and the 10-lb block D. The cord is fixed to a pin at A and passes over two small pulleys at B and C. Determine the weight of the suspended block E if the system is in equilibrium when s = 1.5 ft.
[A].
W = 8.01 lb
[B].
W = 14.91 lb
[C].
W = 17.63 lb
[D].
W = 18.33 lb

Answer: Option D

Explanation:

No answer description available for this question.

Read More A “scale” is constructed with a 4-ft-long cord and the 10-lb block D. The cord is fixed to a pin at A and passes over two small pulleys at B and C. Determine the weight of the suspended block E if the system is in equilibrium when s = 1.5 ft.
Engineering Mechanics | Equilibrium Of A Particle

Determine the magnitude and direction of the resultant force FAB exerted along link AB by the tractive apparatus shown. The suspended mass is 10 kg. Neglect the size of the pulley at A.

Question: Determine the magnitude and direction of the resultant force FAB exerted along link AB by the tractive apparatus shown. The suspended mass is 10 kg. Neglect the size of the pulley at A.
[A].
FAB = 170.0 N, = 75.0°
[B].
FAB = 170.0 N, = 15.0°
[C].
FAB = 98.1 N, = 75.0°
[D].
FAB = 98.1 N, = 15.0°

Answer: Option D

Explanation:

No answer description available for this question.

Read More Determine the magnitude and direction of the resultant force FAB exerted along link AB by the tractive apparatus shown. The suspended mass is 10 kg. Neglect the size of the pulley at A.