Calculate Acceleration and Tension in a 2-Crate System | General Dynamics

[Eric [Tsu]]

1. Two packing crates of masses m1 = 10.0 kg and m2 = 7.00 kg are connected by a light string that passes over a frictionless pulley as in Figure P4.26. The 7.00 kg crate lies on a smooth incline of angle 43.0°.

(a) Find the acceleration of the 7.00 kg crate.
(b) Find the tension in the string.



Drew a diagram identifying all the forces. I know I need to make the equation for m1 and m2, and add them together in order to cancel out all but 1 variable. Unfortunately I've forgotten too many steps to make a decent attempt at that.

Thanks in advance

 

[eok20]

Why don't you post the equations of motion and your attempt at solving them for the acceleration. Did you use the fact that the tension at both ends of the string are the same?

 

[Eric [Tsu]]

Because I'm completely lost.

Edit: Does the normal force equal w2y?

Edit2: My best attempt at m2s x-axis equation:

T-w2x = (m2)a
T-(68.6sin43) = (m2)a

m1s y-axis equation:

T+w1 = (m1)a
T+98 = (m1)a

I know I'm missing something here, I just don't know what

 

[eok20]

That looks good (assuming x is in the direction of the plane). You're right about the normal force too but you shouldn't have to worry about it in this problem since there is no friction and the motion is perpendicular to the normal force.

Your equation for m1 is wrong because you have the tension and the weight in the same direction, but they are opposed to each other. To be consistent with the direction you chose for m2, you should have w1 - T instead of T + w1.

 

[Eric [Tsu]]

Got acceleration. In addition to over-thinking it, I forgot a negative sign as you pointed out.

(-T) - (68.6sin43) = (m2)a
+ T + 98 = (m1)a
_______________________
-(68.6sin43) + 98 = (m2)a + (m1)a

m1 = 10
m2 = 7
Solve for a
--> a = 3.01

Edit: Tension = 67.9 N

T + 98 = (m1)a
T + 98 = (10)3.01
T = 67.9 N

 

[eok20]

What you had before was correct. The netforce on m2 is T - 68.6sin(43). But instead of T + 98 it should be 9.8 - T since the two forces oppose each other.

 

[Eric [Tsu]]

I see, thanks.

Assuming I had to deal with friction, how would I apply that?

 

[Eric [Tsu]]

Here's another without friction:

A mass, m1 = 4.00 kg, resting on a frictionless horizontal table is connected to a cable that passes over a pulley and then is fastened to a hanging mass, m2 = 10.0 kg, as in Figure P4.30. Find the acceleration of each mass and the tension in the cable.

Figure P4.30: webassign.net/sf5/p4_25.gif

Eq for m1x: T-(m1g) = m1a
Eq for m2y: -T+(m2g) = m2a

Added together: -(m1g)+(m2g) = a(m1+m2)

Solved for a: a = 4.2

Where did I go wrong?

 

[eok20]

Friction would oppose the motion of m2 and would have magnitude (coefficient of kinetic friction)*(normal force), the normal force being w2y. Thus if we call the coefficient of friction \mu then the equation of motion for m2 would become T - w2x - \muw2y = m2a if T > w2x and T - w2x + \muw2y = m2a if T < w2x

Note that this assumes that |T-w2x| is bigger than the static frictional force.

 

[alphysicist]

Hi Eric,

Here's another without friction:

A mass, m1 = 4.00 kg, resting on a frictionless horizontal table is connected to a cable that passes over a pulley and then is fastened to a hanging mass, m2 = 10.0 kg, as in Figure P4.30. Find the acceleration of each mass and the tension in the cable.

Figure P4.30: webassign.net/sf5/p4_25.gif

Eq for m1x: T-(m1g) = m1a

For the x equation, you need the x component of the weight. What is that for m1? Once you have that, you should get the answer.

Eq for m2y: -T+(m2g) = m2a

Added together: -(m1g)+(m2g) = a(m1+m2)

Solved for a: a = 4.2

Where did I go wrong?