Finding tensions from incline

In summary: The equation you need to solve for the suspended mass M is Fx = -GMx/L, where F is the force, M is the mass of the block, x is the displacement, and L is the length of the light string.
  • #1
Blocks 1 and 2 of masses ml and m2, respectively, are connected by a light string, as shown above. These blocks are further connected to a block of mass M by another light string that passes over a pulley of negligible mass and friction. Blocks l and 2 move with a constant velocity v down the inclined plane, which makes an angle q with the horizontal. The kinetic frictional force on block 1 is f and that on block 2 is 2f.

I know that the coefficient of friction would be Ff/m1(-9.8)cos(angle).
How would I find the value of the suspended mass M that allows block 1 & 2 to move with constant velocity?
 

Attachments

  • Dynamics AP Problems(2).docx
    130.7 KB · Views: 268
Physics news on Phys.org
  • #2
gymnast2121 said:
I know that the coefficient of friction would be Ff/m1(-9.8)cos(angle).
How would I find the value of the suspended mass M that allows block 1 & 2 to move with constant velocity?
You won't need to consider the coefficient of friction to answer this part.
Do you understand how to develop free body diagrams when there are multiple interacting bodies? You create unknowns for all of the forces of interaction (the tensions here), draw a free body diagram for each body, and write down the dynamical equations for each body. You also need to make use of kinematic facts, like the fact that some string length does not change. This allows you to deduce that some velocities or accelerations are the same. Where a body has no mass (the pulley here) you know the sum of the forces and sum of torques are zero.
 
  • #3
So to find the suspended mass M would I create a free body diagram showing all of the forces acting on it then find what equation to find it? Would the Equation involve using sin or cos with the angle?
 
  • #4
gymnast2121 said:
So to find the suspended mass M would I create a free body diagram showing all of the forces acting on it
That alone won't tell you the mass. You need a free body diagram for each object in the system, and equations obtained from them. When the number of equations matches the number of unknowns, you probably have enough information to solve it.
Since the pulley is massless and frictionless, we can avoid an FBD for that; you just have to figure out how the tension on one side of it relates to the tension on the other side. So three FBDs should do it.
In general, you can get three equations per FBD: two for linear forces along some chosen co-ordinates (often, horizontal and vertical) and one for torque. In the present case there are no torques. So you could get up to 6 equations, but you won't need them all.

What equation do you know relating forces and accelerations in general?
 
  • #5


To find the value of the suspended mass M that allows blocks 1 and 2 to move with constant velocity, we can use the concept of equilibrium. Since the blocks are moving at a constant velocity, we can assume that the net force acting on them is zero. This means that the forces acting in the horizontal direction must be equal to the forces acting in the vertical direction.

In the horizontal direction, there are two forces acting on block 1: the tension in the string (T) and the frictional force (f). Similarly, there are two forces acting on block 2: the tension in the string (T) and the frictional force (2f). The tension in the string (T) is the same for both blocks since they are connected by the same string.

In the vertical direction, there are three forces acting on block 1: the weight of the block (m1g), the normal force (N) from the inclined plane, and the component of the tension in the string (Tsinθ) acting in the upward direction. Similarly, there are three forces acting on block 2: the weight of the block (m2g), the normal force (N) from the inclined plane, and the component of the tension in the string (Tsinθ) acting in the downward direction.

Since the blocks are moving at a constant velocity, we can equate the horizontal forces and the vertical forces to zero. This gives us the following equations:

Horizontal forces:

T + f = T + 2f
f = 2f

Vertical forces:

m1g + N + Tsinθ = m2g + N + Tsinθ
m1g = m2g

From these equations, we can see that the mass of block 1 (m1) is equal to the mass of block 2 (m2). Therefore, the value of the suspended mass M does not affect the constant velocity of blocks 1 and 2. As long as the masses of blocks 1 and 2 are equal, the system will remain in equilibrium and move at a constant velocity down the inclined plane.
 

1. How do I calculate the tension in a string/rope using an incline?

To calculate the tension in a string or rope on an incline, you will need to use the equation T = mg sinθ, where T is the tension, m is the mass of the object, g is the acceleration due to gravity (9.8 m/s^2), and θ is the angle of incline.

2. Can I use the same equation to find the tension in a cable on an incline?

Yes, the same equation T = mg sinθ can be used to calculate the tension in a cable on an incline, as long as the cable is considered to be massless.

3. What if there are multiple objects connected by a string on an incline?

In this case, you will need to consider the tension in the string for each object separately. You can use the same equation T = mg sinθ for each object, but plug in the individual masses and angles of incline for each one.

4. Is it possible to have a negative tension on an incline?

No, tension is a force and forces are always positive. If the calculated tension comes out to be negative, this means that the direction of the tension force is opposite to the assumed direction. You can simply change the direction in your calculation to get a positive value.

5. How does the angle of incline affect the tension?

The tension in a string or rope on an incline will increase as the angle of incline increases. This is because the component of the weight force acting downwards along the incline also increases, resulting in a larger tension force to counteract it.

Suggested for: Finding tensions from incline

Replies
28
Views
246
Replies
2
Views
575
Replies
2
Views
349
Replies
4
Views
2K
Replies
10
Views
945
Replies
10
Views
1K
Replies
6
Views
965
Replies
7
Views
2K
Back
Top