See our solution for Question 41P from Chapter 16 from Hibbeler's Engineering Mechanics.
Problem
Chapter: 16
Problem: 41P
At the instant theta = 50 degrees, the slotted guide is moving upward...
Step-by-Step Solution
Engineering
Mechanics:
Statics and
Dynamics, 14th
Edition
Authors: Russell C.
Hibbeler
ISBN-13: 9780133915426
Step 1
We are given a slotted guide that is moving in the upward direction.
The angle of the slotted guide from the vertical is, θ
The acceleration of the slotted guide is, a
The velocity of the slotted guide is, v
= 3 m/s
= 50
∘
.
2
= 2 m/s
We are asked to determine the angular acceleration and angular velocity of link AB.
Step 2
The free-body diagram of the slotted guide is given by,
Here, ω is the angular velocity and α is the angular acceleration.
Step 3
From the right-angled triangle formula, the vertical height of the link AB is given by,
y = (0.3 m) cosθ
On differentiating the above expression with respect to time, we get,
dy
v =
v =
d
dt
dt
((0.3 m) cosθ)
˙
v = − (0.3 m) sin θ (θ )
… (1)
On substituting v
,
= −2 m/s θ = 50
∘
and θ˙
= ω
in equation (1), we get,
∘
− (2 m/s) = − (0.3 m) (sin 50 ) (ω)
(2 m/s)
ω =
(0.229 m)
ω = 8.703 rad/s
Step 4
On differentiating the equation (1) with respect to time, we get,
a =
a =
d
dvy
dt
˙
((−0.3 m) sin θ (θ ))
dt
2
¨
˙
a = −0.3 (sin θ ⋅ θ + cos θ ⋅ θ )
On substituting a
= −3 m/s
2
,θ
= 50
2
∘
, θ˙
= ω
(−3 m/s ) = −0.3 m (sin 50
sin 50
∘
∘
and θ¨
= α
in the above expression, we get,
⋅ (α) + cos 50
∘
⋅ (α) = −38.686rad/s
α = −50.50rad/s
2
2
⋅ (8.703rad/s) )
2