EGR 280 – Mechanics Exam
Statics of particles and rigid bodies
∑F = R = 0and∑MO = 0MO = rO×F
g = 9.81 m/s2 = 32.2 ft/s2
Normal stress, deformation
σ = P/Aε = δ/Lσ = Eεδ = PL / AE
Friction
static equilibrium: f ≤ μsN ; if moving, f = μkN
belt friction: T2 = T1 eμβ
Particle kinematics
v = dx/dta = dv/dt = d2x/dt2 = v(dv/dx)
uniform rectilinear motion: x = x0 + v0t
uniformly accelerated rectilinear motion:
v = v0 + a0tx = x0 + v0t + ½ a0t2v2 = v02 + 2a0(x-x0)
relative motion: xB = xA + xB/AvB = vA+ vB/AaB = aA + aB/A
intrinsic coordinates: v = v eta = dv/dt et + v2/ρ en
Particle kinetics
Newton’s Second Law:∑F = ma
intrinsic coordinates:∑Ft = m dv/dt∑Fn = m v2/ρ
work and energy:TA + UAB = TB
kinetic energy:T = ½mv2
work done by spring:U = ½k δ2 ; by weight: -Wy
conservation of energy:VA + TA = VB + TB
if not conserved, then:VA + TA = VB + TB + energy lost
impact and momentum:mv1 + ∫Fdt = mv2
coefficient of restitution:e = (v´B-v´A)/(vA-vB) = v´B/A/vA/B
Rigid body kinematics – plane motion
translation:vA = vBaA = aB
rotation:v = ωk × r a =αk × r – ω2r
uniform rotation: θ = θ0 + ω0t
uniformly accelerated rotation:
ω = ω0 + α0tθ = θ0 + ω0t + ½ α0t2ω2 = ω02 + 2α0(θ-θ0)
general plane motion:vB = vA +ωk × rB/Aa= aA + αk × rB/A – ω2rB/A
rolling:v = ωra = αr
pulleys:ω1r1= ω2r2gears: ω1r1= -ω2r2
mass moments of inertia:
parallel axis theorem: IO = IG + md2 ; radius of gyration: k2 = I/m
rod: IG = mL2 / 12disk: IG = mr2 / 2
ring: IG = mr2sphere: IG = 2mr2 / 5
Rigid body kinetics – plane motion
Newton’s Second Law: ∑F = ma∑MG = IGαk
non-centroidal rotation: ∑Mo = Ioαk
kinetic energy:T = ½mvG2 + ½ IGω2 ; non-centroidal rotation: T = ½ IOω2
total momentum:mvG + IGω ; non-centroidal rotation: Ioω
- Given force F = (3i + 5j + 4k)lb passing through point (4,12,6)ft, find the y-component of the moment of F about the origin.
- -2 ft-lb
- 0
- 2 ft-lb
- 14 ft-lb
- Given F = (3i + 5j + 4k)N and r = (6i – 3j + 12k)m, find F∙r.
- -21 N-m
- (-72i + 12j + 39k)N-m
- 42 N-m
- 51 N-m
- The block shown is acted upon by a force P = 7 lb. If μs=0.8 and μk=0.5, determine the acceleration of the block.
- 0
- 3.22 m/s2
- 6.44 m/s2
- 25.8 m/s2
- Determine the force in member AB.
- 900 N (T)
- 900 N (C)
- 1200 N (T)
- 1200 N (C)
- The reaction at B is:
- 3.5 lb ↑
- 3.5 lb ↓
- 16.5 lb ↑
- 16.5 lb ↓
- Find the force in member BD of the simple truss.
- 750 N (C)
- 1000 N (C)
- 1000 N (T)
- 3000 N (C)
- A block will slide down a 30° incline if the coefficient of static friction is less than
- 0.500
- 0.577
- 0.866
- 0.625
- What force F is required for equilibrium?
- W / 4
- W / 6
- W / 8
- W / 9
- One complete revolution of a rope about a post is used to hold a boat. If μs=0.5, what maximum force by the boat can be resisted if the holder provides a force of 100N?
- 962 N
- 1920 N
- 2310 N
- 3850 N
- A 30-mm diameter rod supports a tensile force of 35 kN. What is the normal stress in the rod?
- 12.4 MPa
- 36.4 MPa
- 42.8 MPa
- 49.5 MPa
- A weight of 40 kN is suspended from a 20-m long, 20-mm diameter steel cable. If E = 207 GPa, determine the elongation of the cable.
- 3.1 mm
- 8.7 mm
- 10.8 mm
- 12.3 mm
- The Work and Energy principle is a relationship between
- Force, acceleration and time
- Force, speed and distance
- Force, velocity and time
- All of the above
- Impulse and Momentum is a relationship between
- Force, acceleration and time
- Force, speed and distance
- Force, velocity and time
- All of the above
- A ball is given an initial velocity of 40 m/s straight up. Ignoring friction, how high will the ball go?
- 40 m
- 60.8 m
- 81.6 m
- 122 m
- A wheel rotates at 20 rad/s. How many revolutions will it rotate in 4 s, after it begins to decelerate at 10 rad/s2?
- 3.18 rev
- 4.25 rev
- 6.37 rev
- 7.40 rev
- A car is traveling at a constant speed around a circular curve having a radius of curvature of ρ = 250 m. If the magnitude of the car’s acceleration is 1.5 m/s2, determine the speed at which the car is traveling.
- 19.4 m/s
- 23.7 m/s
- 32.8 m/s
- 34.1 m/s
- The center of the wheel is moving to the right with a speed of 2 m/s. If no slipping occurs at the ground, determine the velocity of point B at the instant shown.
- 1.33 rad/s cw
- 4 m/s →
- 3 m/s →
- 2 m/s →
- If the system shown is released from rest, find the tension in the cord which is wrapped around the 50-kg cylinder.
- 80 N
- 99 N
- 100 N
- 109 N
- The 20-kg mass in Problem 18 drops from rest. If it falls through 2 m, what is the angular speed of the cylinder?
- 8.4 rad/s
- 9.3 rad/s
- 10.4 rad/s
- 12.2 rad/s
- The 35-kg block strikes the unstretched spring. The maximum deflection of the spring is:
- 1.67 m
- 2.80 m
- 4.78 m
- 6.33 m
- The 6-lb ball is fired from rest using a spring as shown. Determine how far the spring must be compressed so that when the ball reaches a height of 8 ft it has a speed of 6 ft/s.
- 0.80 ft
- 1.27 ft
- 1.60 ft
- 2.56 ft
- A ball is dropped vertically downward onto a horizontal surface from a height of 52 m. If the coefficient of restitution e = 0.8, determine the speed of the ball after it bounces.
- 20.5 m/s
- 25.6 m/s
- 30.6 m/s
- 32.0 m/s
- The clock pendulum consists of a 1-kg slender rod A and a 4-kg moveable disk B. Determine the mass moment of inertia of the pendulum about the fixed point O.
- 0.10 kg-m2
- 1.97 kg-m2
- 2.30 kg-m2
- 3.14kg-m2
- In the belt and pulley system shown, if pulley Ais rotating at 1740 rpm, what is the angular speed of pulley D? Pulley C is attached to and rotates with pulley B.
- 196 rpm
- 870 rpm
- 1390 rpm
- 2180 rpm
- In the gear system shown, if gear A is rotating at 1740 rpm clockwise, determine the angular velocity of gear E. Gear B is attached to and rotates with gear C.
- 61.9 rpm ccw
- 61.9 rpm cw
- 96.7 rpm ccw
- 96.7 rpm cw