M1 Problem

Two bricks P and Q each of mass 5kg, are connected by a light inextensible string. Brick P is held at rest and Q hangs freely, vertically below P. A force of 200N is then applied vertically upwards to P causing it to accelerate at 1.2ms^-2. Assuming there is a resistance to the motion of each of the bricks f magnitude RN, find

a) The Value of R
b) The tension in the string connecting the bricks.

For part a I've drawn the diagram and used
R=ma+mg
R=(5*1.2)+(5*9.8)
Which gave me 55N, however, the book states that answer is 45N

Here is my diagram

Any ideas on what I have done wrong? :smile:

Thank you!

(edited 9 years ago)

Original post by Xenosa

Two bricks P and Q each of mass 5kg, are connected by a light inextensible string. Brick P is held at rest and Q hangs freely, vertically below P. A force of 200N is then applied vertically upwards to P causing it to accelerate at 1.2ms^-2. Assuming there is a resistance to the motion of each of the bricks f magnitude RN, find

a) The Value of R
b) The tension in the string connecting the bricks.

For part a I've drawn the diagram and used
R=ma+mg
R=(5*1.2)+(5*9.8)
Which gave me 55N, however, the book states that answer is 45N

Here is my diagram

Any ideas on what I have done wrong? :smile:

Thank you!

Moved to Maths section :smile:

Reply 2

ghostwalker

17

Original post by Xenosa

Any ideas on what I have done wrong? :smile:

Thank you!

Two things:

There is a resistance of RN to each brick - i.e. 2R N in total.

For part a) you need to treat the two masses as one. The 200N force although applied to P, causes both P and Q to accelerate at the same velocity, as they're joined by a taut string.

Reply 3

waxwing

5

In problems like this, involving composites of more than one object, it's important to be clear what you are drawing a free body diagram of. You can choose to analyse the motion of (and draw a free body diagram of) either the upper mass, the lower mass, or the whole system.

Note that if you choose to apply Newton's second law to the whole system, then the resistance force will be

2R

.

Another note: to see why it's a good idea to apply Newton's second law to the whole system, note that in this case the Tension forces are internal to the system (forces between parts of the system), and thus cancel - whereas the resistance forces are not internal.

Reply 4

Xenosa

OP

Original post by ghostwalker

Two things:

There is a resistance of RN to each brick - i.e. 2R N in total.

For part a) you need to treat the two masses as one. The 200N force although applied to P, causes both P and Q to accelerate at the same velocity, as they're joined by a taut string.