Mech2.
m1
mb
m2
A mb=15kg ball hangs by a string from the roof of a m1=20kg box that is attached to a m2=10kg rock over
a light frictionless pulley. There is no appreciable friction between the box and the table it rests on.
a) Draw a free-body diagram of the ball as the system is moving(below) (3points)
Sol) There can be two ways to draw the free-body diagram. In perspective to an observer inside the box,
and to an observer outside the box. Both are correct.
i) For an observer inside the box, a force called fictitious force
ii) For an observer outside the box,
exists due to the acceleration of the box. But since the acceleration the ball will move with the same
of the box is constant, the net force acting on the ball is zero. So
acceleration as the box
the hanging ball will ‘seem’ to be stationary but at the angle θ
a
T’cosθ
T’
T’cosθ
T’
Ffictitious
T’sinθ
T’sinθ
mbg
mbg
b) Find the acceleration of m1 (4points)
Sol) m1 : ∑Fx = T = (m1+mball)a
m2 : ∑Fy = m2g - T = m2a
*~ T is the tension force acting on m1 and T’ is the tension force acting on mb
Solving for 'a' gives
a=𝑚
𝑚2 𝑔
1 +𝑚𝑏𝑎𝑙𝑙 +𝑚2
= 2.18m/s2
c) What angle(θ) does the string inside the box make with the vertical while motion is occurring?
(4points)
Sol1) For an observer inside the box
Sol2) For an observer outside the box
∑Fx=T’sinθ – Ffictitious =0 , ∑Fy=T’cosθ - mballg=0 ∑Fx = T’sinθ = mballa , ∑Fy = T’cosθ - mballg=0
where Ffictitious =mballa
If we divide the two above equations
If we divide the two above equations
tanθ =a/g
tanθ =a/g
So,
So,
θ=tan-1(0.222) = 12.5°
θ=tan-1(0.222) = 12.5°
d) If you could vary the weight of the rock, what is the maximum angle the string could make with
the vertical? (4points)
Sol) When the mass of the rock is large relatively to the box and ball, then the rock will nearly free-fall.
At this point, the ball will have the maximum angle.
When m2>>m1 and m2>>mball, then
a=𝑚
𝑚2 𝑔
1 +𝑚𝑏𝑎𝑙𝑙 +𝑚2
= 𝑚1
𝑚2
𝑔
𝑚
+ 𝑏𝑎𝑙𝑙 +1
≈g
𝑚2
So if a=g, then
tanθ =g/g=1
θ=45°
Mech3.
m
4m
m
In the figure above, the top block is one-fourth the length of the bottom block and weighs only one-fourth
as much. Assume that there is no friction between the bottom block and the surface on which it moves
and the coefficient of kinetic friction between the two blocks is μk =0.2.
a) Find the acceleration of top block m. (4points)
Sol) The acceleration of the top block and hanging
block is a1.
Top m; ∑Fx = T- fk = ma1 , ∑Fy = FN - mg=0
FN
m
T
fk
Hanging m ; ∑Fy = mg - T=ma1
Eliminating T will gives, mg - μkmg=2ma
m
Fg
=> a1=3.92m/s2
b) Find the acceleration of bottom block 4m. (3points)
Sol) Only fk acts on block 4m and since the surface is
4m
frictionless, we do need to use the FN.
∑Fx = fk = 4ma2 , where is the acceleration of 4m
Since fk = μkmg, a2=0.49m/s2
fk
x1
x2
(c) The length of the bottom block is ‘L’. After the system is released, find the distance(x1) the top
block has moved when only one-fourth of the top block is still on the bottom block. The distance
must be expressed in term of L only. (4points)
3
3
Sol) x1 = x2 + 4 𝐿 + 16 𝐿
= x2 +
15
𝐿
16
The time it takes for the top block to travel x1 is equal to the time it takes for the bottom block to travel x2.
Inserting the values a1=3.92m/s2 and a2=0.49m/s2
1
1
x1 = 2a1t2 = 1.96t2 , x2 = 2a2t2 = 0.245t2
15
15
Since x1 = x2 + 𝐿, 1.96t2 = 0.245t2+ 𝐿. So t2 =0.547L.
16
16
=> x1=1.07L
(d) Find the distance(x2) the bottom block has moved in term of L only. (4points)
1
Sol) x2= a2t2 = 0.245 ×0.547L = 0.134L
2