Force (object is suspended connected to a block on table and pulled up)

[~christina~]

Homework Statement

A block of mass m1 on a rough horizontal surface is connected to a ball of mass m2 by a lightweight cord over a lightweight frictionless pulley as shown. A force of magnitude F at an angle theta with the horizontal is applied to the block. The coefficient of friction between the block and surface is \mu_k

a.) draw a free body diagram of the 2 masses

b.) use Newton's dynamic equation to determine the magnitude of the acceleration of theh 2 objects in terms of their masses, F, theta, and \mu_k

c.) what is the tension in the cord
http://img518.imageshack.us/img518/1023/picture1dn4.png

Homework Equations

F= ma

Ffk= uk* Fn

and a Newton's dynamic equation? ==> what is that exactly??

The Attempt at a Solution

a)for the free body diagram

http://img484.imageshack.us/img484/2900/90134448jw8.th.jpg

b) how would I do this question...not sure and I'm stuck after the free body diagram...hm..


c) I guess I would need to solve part b to get part c


~I do know that they both have the same accelration but how would I know if the total forces \sum Fx= 0?

I need help please..

 

[Doc Al]

and a Newton's dynamic equation? ==> what is that exactly??

That's Newton's 2nd law.

One problem with the FBD: That applied force F is at an angle, not horizontal.

Analyze the forces parallel to the direction of motion for each block and apply Newton's 2nd law. You'll get two equations--one for each mass--that you'll solve together for the two unknowns: the acceleration and tension.

 

[~christina~]

I was wondering why I don't analyze the horizontal component of the block but I think it's b/c the total forces = 0 but...this is what I got.

for m2 (only movement in y direction)
\sum Fx= 0
\sum Fy= T- m_2 g= m_2 ay

for m1 (block)
\sum Fx= max= Fcos theta- (T+ \muk m1g)
\sum Fy= may= Fsin theta + N- mg= 0

Well I just wanted to check first...

 

[Doc Al]

for m2 (only movement in y direction)
\sum Fx= 0
\sum Fy= T- m_2 g= m_2 ay

Good. I would just call the acceleration "a".

for m1 (block)
\sum Fx= max= Fcos theta- (T+ \mu_k m1g)

Careful here: What's the normal force? Again: call the acceleration "a".

\sum Fy= may= Fsin theta + N- mg= 0

Good. Use this to figure out N.

 

[~christina~]

for m2 (only movement in y direction)
\sum Fx= 0
\sum Fy= T- m_2 g= m_2 a

for m1 (block)
\sum Fx= ma= Fcos theta- (T+ \muk m1g) => was this incorrect (see below for my correction)??

~I do know that that is the Fx component right? but I guess I need to incorperate the N?


\sum Fy= ma= Fsin theta + N- mg= 0
from the Fy

N= mg - Fsin theta


I think that since the Ffk = uk*FN
then the equation on the top for the Fx wouldn't be uk* m1g since the normal force is lighter so it would be instead

\sum Fx= ma= Fcos theta- (T+u_k (m_1g- Fsin theta))


Is it fine now?

I guess whenever I have to find the x component of the force when there is friction I need to incorperate the normal force the angle it is being pulled at if there is a angle so that the normal force is not just m*g right?

 

[Doc Al]

Is it fine now?

Yes, all better.

I guess whenever I have to find the x component of the force when there is friction I need to incorperate the normal force the angle it is being pulled at if there is a angle so that the normal force is not just m*g right?

Right. Friction depends on the normal force, which may or may not equal the weight of the object, depending upon the situation.

 

[~christina~]

\sum Fy= T- m_2 g= m_2 a

\sum Fy= ma= Fsin theta + N- mg= 0
from the Fy

N= mg - Fsin theta

\sum Fx= ma= Fcos theta- (T+u_k (m_1g- Fsin theta))

________________________________________________________________
but how would I find the part b) which is...

b) b.) use Newton's dynamic equation to determine the magnitude of the acceleration of the 2 objects in terms of their masses, F, theta, and \mu_k

(I'm not sure what they want)

would it be adding the 2 equation??

the first object and second object's forces like this?

\sum Fy= T- m_2 g= m_2 a

\sum Fx= ma= Fcos theta- (T+u_k (m_1g- Fsin theta))

so adding them I get..
+ T- mg= m2a
+ Fcos theta - T- uk (m1g- Fsin theta) = m1a
______________________________________

Fcos theta- m2g- uk(m2g- Fsin theta)= (m1 + m2)a

a= (Fcos theta- m2g - uk(m2g- Fsin theta))/ m1+ m2
Is this what they want?


Assuming it is...
for part c.) I need to find

c.) tension in cord..

I think that I would plug this into a original equation while using the acceleration I found...

\sum Fy= T- m_2 g= m_2 a

T= m2a + m2g

substituting a

T= m2( (Fcos theta- m2g - uk(m2g- Fsin theta))/ m1+ m2) + m2g


I think this is okay but not sure..

 

[Doc Al]

Looks good to me!

As a practical matter, I would solve for a and then plug the numerical value into the other equation to get T.

 

[~christina~]

Oh okay..I would plug in the numbers if I had numbers given...

I usually find it challenging to not have numbers though..

Thanks for your help Doc Al !