Interacting Systems - The Sled Dog!

1. Oct 17, 2007

mantillab

1. The problem statement, all variables and given/known data
The sled dog in figure (attached) drags sleds A and B across the snow. The coefficient of friction between the sleds and the snow is 0.10.

If the tension in rope 1 is 150 N, what is the tension in rope 2?

2. Relevant equations

F_a on b = -F_b on a

3. The attempt at a solution
The force diagrams I did for sled A and sled B resulted in:

Sled A: Fnetx = T1 - (mu_k)(m_a)(g) = 52

Sled B: Fnetx = T2 -T1 - (mu_k)(m_b)(g) = T2 - 228.4

My solution for T2 = 280 is very close, but not the right solution. Should I be adding another force for sled B, based on the friction of sled A or another component of sled A?

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2. Oct 17, 2007

elyes

can you help me please solving my problem thx

3. Oct 17, 2007

mantillab

Apparently the answer is 270N. Still don't know they determined that.

4. Oct 17, 2007

What are the masses of the sleds? I can't see the picture but I'm imagining that a dog is pulling sled b which is connected to sled a like so.

Dog--b--a

I don't know if this is what you meant, but Fnet for sled a and b are NOT the same unless they have the same mass.

5. Nov 4, 2007

Kster

Solution

T for rope 1:
100kg * a = 150 N
a = 1.5 m/s^2

If the tension in rope 1 is 150 N, what is the tension in rope 2?

T for rope 2:
(mass of sled 1 + sled 2)*1.5 m/s^2 = T
(100kg + 80kg)*1.5m/s^2 = 270N

6. Oct 26, 2009

xstetsonx

kster you ignored the friction force.
Rope 1:
T-Mkn=ma
150-0.10(100*9.8)=ma
rope 2 just the same expect for the weight is 180kg

7. Oct 30, 2009

damionpitt

So to Sum it all up...

Remember when dealing with any type of friction problem, you MUST add it into the eq. usually by finding the F_net along the y-axis.

F_net_y = m_1 *a_y (a_y = 0, since the object is not moving in the y-direction)

Find all forces acting on the y-axis to be your F_net_y

F_net_y = N - m*g
N - m *g = m_1 *a_y
N - m *g = 0
N = m *g

Also remember what your Friction Force is.

F_k = U_k * N
F_k = U_k * (m *g)

Now you can solve for a

T_1 - F_k = m_1 *a
(T_1 - F_k)/(m_1) = a
(T_1 - (U_k *(m_1 *g)))/(m_1) = a

Then you solve for T_2

T_2 - T_1 - F_k = m_2 * a
T_2 = (m_2 * a) + T_1 + F_k
T_2 = (m_2 * a) + T_1 + (U_k *( m_2 *g))