Engr 214 Quiz 3 Solutions
1.) The constraint forces do no work so only gravity need be considered in the work energy equation.
The potential energy of the cart is constant so the total potential energy of the system may be taken
as V = mB gyB , where yB i
Engr 214 Quiz 3
April 9, 2003. Closed Books and Notes
Class supplied equation sheet permitted.
Instructions: Solve any 2 of the three problems below. Present your answers clearly, e.g., by
enclosing them in a box.
1.) A 20 kg mass B is suspended from a 2
Engr 214 Quiz 2 Solutions
1.) Let xA be the position of block A measured from the cable tie down (increasing to the left), and
xB the position of block B from the same point (increasing to the right). The equations of motion
for the two blocks are: mA xA
Engr 214 Quiz 2
March 19, 2003. Closed Books and Notes
Class supplied equation sheet permitted.
1.) Neglecting the eect of friction, determine (a) the acceleration of each block, (b) the tension in
the cable.
125 N
A
B
25 kg
25 kg
2.) A space shuttle is d
Engr 214 Quiz 1
February 21, 2003. Closed Books and Notes
Class supplied equation sheet permitted.
Instructions: Solve problem 1 and either problem 2 or 3. Present your answers clearly, e.g., by
enclosing them in a box.
1.) A particle moves in the x, y pl
Engr 214 Quiz 1 Solutions
1.) a. The velocity vector is v = (2t+3)+8(2tt2 )et and the acceleration a = 2+8(24t+t2 )et .
At t = 0.5s, v = 4 + 6e0.5 = 4 + 3.64 m/s, and a = 2 + 2e0.5 = 2 + 1.21 m/s2 .
b. The speed v = v = 16 + 13.25 = 5.41 m/s
c. The tangen
Engr 214 Assignment 11 Solutions
Note: The solution to all the problems below is based on the dAlemberts principle which can be
formulated as follows: Dene the inertia force on a rigid body B by the equation F = maG ,
where m is the total mass of B , aG i
Engr 214 Assignment 10 Solutions
15.10) The angular velocity of the body is = n, where n = (20 + 9 12k )/25 = 0.8
0.36 + 0.48k , and = 7.5 rad/s. The velocity of point E is (use the formula relating the velocity of
two points on the same rigid body) vE =
Engr 214 Assignment 9 Solutions
14.7) During the interaction of the bullet with block A x-momentum is conserved (x is the direction
of travel of the bullet). Thus, mb vb = mA vA + mvb where vb = 1500 ft/s is the initial velocity of
the bullet, vb its velo
Engr 214 Assignment 8 Solutions
13.137) Using Fig. P137, the magnitude of the applied force is given by
At, A = 5lb/s, 0 t 2s
P (t) = 2A, 2 < t 3s
0,
t > 3.
While the collar is in contact with the support (and hence at rest), the total force on the colla
Engr 214 Assignment 6 Solutions
12.95) The path of the particle is r = r0 sin . We know that u = 1/r() satises the equation (12.37)
u + u = F/(mh2 u2 ) with F (r) the magnitude of the force on the particle. Dierentiating we nd:
u = cos /(r0 sin2 ), and u
Engr 214 Assignment 5 Solutions
12.72) Let r be the distance of the slider from the axis of rotation, and the angle between a xed
line in the (horizontal) plane of the plate and the slot in which the slider moves. The horizontal
components of the equation
Engr 214 Assignment 4 Solutions
12.6) We are given that the total vehicle weight W is distributed so that in a front-wheel drive
conguration the driving wheels support the fraction f = 0.62 of the total weight. In this case, the
maximum horizontal force t
Engr 214 Assignment 2 Solutions
11.39) We are given that the position x(t) of the police car satises: x = a, x(0) = x(0) = 0 with
a constant. Then v = x = at so that the information v = 90000/3600 = 25 m/s at t = 8s implies
a = 25/8 m/s2 . Thus, at t = 8s
Assignment 1 Solutions
11.5) Given: x(t) = 6t4 2t3 12t2 + 3t + 3. The acceleration is a = x = 72t2 12t 24. a = 0
6t2 t 2 = (3t 2)(2t +1) = 0. Thus, t = 2/3s. Substitute into x(t), v (t) = x(t) = 24t3 6t2 24t +3
to nd the position and velocity.
11.15) Thi