Tutorial Assignment 4 Solutions
APM2EA1
6 March 2017
Tutorial Assignment 4 is worth 10 Marks
Problem 1:
The following figure illustrates the problem:
The forces acting on block A are summed to the null vector (applying the
equilibrium condition):
f + N A + W A + T A = 0,
where T A =: T .
The forces acting on block B are summed to the null vector:
T B + W B = 0,
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where T B = T .
None of the forces acting on block B have non-zero x-components. Applying the equilibrium condition to the y-components of block B results in
−WB + T = 0
⇒ T = WB .
Applying the equilibrium condition to the x-components of block A:
−f + T = 0
⇒ f = T.
Applying the equilibrium condition to the y-components of block A results
in
−WA + NA = 0
⇒ NA = WA .
With f = µNA , then
f =T
⇒ µNA = WB
⇒ µWA = WB
WB
.
⇒µ=
WA
Problem 2:
The following figure defines the tensions:
2
N1 for C:
T 1 + T 2 + W B = 0.
For the x-components:
T1 cos 65◦ − T2 cos 20◦ = 0
⇒ T1 = T2
cos 20◦
cos 65◦
.
For the y-components:
T1 sin 65◦ − T2 sin 20◦ − WB = 0
cos 20◦
⇒ T2
sin 65◦ − T2 sin 20◦ = WB
cos 65◦
⇒ T2 (cos 20◦ sin 65◦ − sin 65◦ cos 20◦ ) = WB cos 65◦
cos 65◦
⇒ T2 = WB
sin 45◦
√
⇔ T2 = 2WB cos 65◦ .
Since the tension in rope F C is the same at all points, then T3 = T2 .
N1 for F :
T 3 + T 4 + W A = 0.
3
For the x-components:
T3 cos 20◦ − T4 cos 40◦ = 0
cos 20◦
⇒ T4 =
.
cos 40◦
For the y-components:
T3 sin 20◦ + T4 sin 40◦ = 0
◦
⇒ WA = T3 sin 20 + T3
cos 20◦
cos 40◦
sin 40◦
T3
(sin 20◦ cos 40◦ + cos 20◦ sin 40◦ )
cos40◦
sin 60◦
= T2
cos 40◦
!
√
3
√
2
= 2WB cos 65◦
cos 40◦
!
√
3 cos 65◦
= √
WB
2 cos 40◦
=
= 67.6 N.
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