Pulley Problem: An object of mass m 1 (= 25 kg) sits on a horizontal frictionles

Question

Pulley Problem:
An object of mass m1 (= 25 kg) sits on a horizontal frictionless table. The rope attached to it runs horizontally to a pulley and down to a hanging mass m2 (= 10 kg).

(Please use the x and y directions shown to answer the questions below.)

a.) Find the acceleration of each mass, and the size of the tension in the rope:
(Note: Make sure to find the general expressions first; this is an extremely important skill. Only after that, plug in the numerical values.)

m1: a1x =  m/s2 ,

a1y = m/s2

m2: a2x =

a2y =

T = ____N, ((Is it less, the same, or more than the weight of the hanging mass? Does this make sense?)

Checking limits: Using your general expression from above,

b.) if m1 was instead zero, what would be the acceleration of m2?

a2x = m/s2

a2y =

c.) if m2 was instead zero, what would be the acceleration of m1?
a1x =

a1y =

d.) With this pulley set-up (and a non-zero m2), is it possible for m1 to remain stationary? What about if the table was not frictionless? Explain.

Explanation / Answer

part a)
considering m1:
T=m1*a
=25*a

considering m2:
m2*g - T= m2*a
10*9.8 - 25*a = 10*a (puting T= 25*a from above equation)
35a = 98
a = 2.8 m/s^2

T=25*a
= 25*2.8
= 70 N

Answers:
m1: a1x = 2.8 m/s2 ,
a1y = 0 m/s2 It is free to move only in x direction
m2: a2x = 0 m/s2 It will be move only in y direction
a2y = 2.8 m/s2

T = 70 N

It is less than weight

Part b)
A.
Let initial velocity be V
then,
Vx = V* cos 65 = 0.423 V
Vy = V* sin 65 = 0.906 V

let time taken be t

Horizontal motion:
d = Vx/t
18 = 0.423V/t
V= 42.55 t
t = 0.0235 V

Vertical motion:
H = Vyi*t - 0.5*9.8*t^2
8 = 0.906V * 0.0235 V - 4.9* (0.0235V)^2
8 = 0.02129 V^2 -0.00271 V^2
8 = 0.01858 V^2
V = 20.75 m/s
Answer: 20.75 m/s

time taken, t = 0.0235 * V = 0.0235 * 20.75 = 0.49 s

b)
If m1 was 0,m2 will fallfreely with acceleration of 9.8m/s^2

a2x = 0
a2y =9.8 m/s^2

c)
If me was zero, m1 will not move at all
a1x = 0
a2y = 0

part d)
No,its not possible.
m1 can remain stationary if there issome other force acting in backward direction on table.

yes if table has friction, that fraction will backward on m1 and it may not allow masses tomove

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