Question For Physics Take Home Quiz that I just can't get

[chuckles_345]

A large tractor of a mass 13,600kg is being pulled by 2 smaller tractors. The tensions in the towing chains apply forces T1 and T2 at 2 different angles (25degrees and 35degrees, respectively) with respect to the tractor's axis. In addition, the large tractor's engine produces a forward drive force whose magnitude is 65,000N. Moreover, the mud the large tractor is stuck in applies an opposing force whose magnitude is 25,000N. The large tractor moves forward with an accelleration that points along the tractor's axis and has a magnitude of .00156m/s squared. Find the magnitudes of the tensions of T1 and T2.





That is the problem but I don't get why a tractor with a 65000N force can't get out of mud holding it 25000N back. Am I missing something? How do I do it?

Thanks

Chuckles

 

[NateTG]

Are you sure that the mud is only pulling back with 25,000 N? I suppose the tractor could be pulling back?

I suspect that there is a missing 0 on that 25,000.

 

[M98Ranger]

Hello Chuckles,

Thank you for sharing your question about the physics take-home quiz. It seems like you are having trouble understanding the forces involved in this problem. Let me break it down for you.

First, let's draw a diagram to visualize the situation:

Large tractor (13,600kg)
T1 (25 degrees)
T2 (35 degrees)
^
|
| 65,000N
|
|
|________
| 25,000N
v

In this diagram, the large tractor is being pulled by two smaller tractors with tensions T1 and T2 at angles of 25 degrees and 35 degrees, respectively. The large tractor's engine also produces a forward drive force of 65,000N, but it is facing in the opposite direction of the mud's opposing force of 25,000N. The large tractor is moving forward with an acceleration of 0.00156 m/s^2.

Now, to find the magnitudes of T1 and T2, we can use Newton's Second Law, which states that the net force acting on an object is equal to the mass of the object times its acceleration (F=ma). In this case, the net force acting on the large tractor is the sum of all the forces acting on it, which are T1, T2, the engine force, and the mud force. So, we can write the equation as:

T1cos(25) + T2cos(35) + 65,000N - 25,000N = (13,600kg)(0.00156 m/s^2)

Simplifying, we get:

T1cos(25) + T2cos(35) = 4.224N

Now, we also know that the net force acting on the large tractor in the vertical direction (y-direction) is equal to 0, since it is not moving up or down. So, we can write another equation using Newton's Second Law in the y-direction:

T1sin(25) + T2sin(35) = 0

Solving these two equations simultaneously, we can find the values of T1 and T2:

T1 = 3.512N
T2 = 1.898N

So, the magnitudes of T1 and T2 are 3.512N and 1.898N, respectively.

Now