Tension problem on an inclined plane

In summary, when the blocks are released, they move due to the tension in the string. The tension in the string is 8.82 Newtons.
  • #1
chronos12
4
0

Homework Statement



Block 1 of mass m1=2.0kg and block 2 of mass m2=3.0kg are connected by a string of negligible mass and are initially held in place. Block 2 is on a frictionless surface tilted at theta = 30 degrees. The coefficient of kinetic friction between block 1 and the horizontal surface is .25 The pulley has negligible mass and friction. Once they are released, the blocks move. What then is the tension in the string?

Please see the attached diagram I drew in paint.

Homework Equations



F = ma (Newtons second law)

The Attempt at a Solution



I worked out a solution but I wanted someone to double check my answer because I've been having problems with this:

T = Tension
fk = kinetic friction

Equation for Mass 1: T - (m1)(g)(fk) = (m1)(a)
Equation for Mass 2: (m2)(g)(cos30) - T = (m2)(a)

Combined: (m1)(a)+(m1)(g)(fk) = (m2)(a)+(m2)(g)(cos30)

Plugging in the masses, g = 9.8 and fk = .25 I come up with a = 4.112

Plugging a back into the equation for mass 1 gives me T = 13.124 N

Thanks for any help. I feel like I'm missing part of it or I need to initially set a to 0?
 

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  • #2
Ηι chronos12, welcome to PF.
Equation for Mass 2: (m2)(g)(cos30) - T = (m2)(a)
Check this equation. In the diagram you have not shown θ.
 
  • #3
Please see the attached revised diagram showing the angle. There should be a negative sign in front of (m2)(a). On paper I wrote this down - just didn't type it correctly here. The combined equations from my work up should read: (m1)(a)+(m1)(g)(fk) = -(m2)(a)+(m2)(g)(cos30).

When you say check equation 2, I'm thinking this should have been sin(30) instead of cos(30)?

It would then read: (m2)(g)(sin30) - T = (m2)(a)

Combined: (m1)(a)+(m1)(g)(fk) = -(m2)(a)+(m2)(g)(sin30)
The new value for a = 1.96
The new value for T = 8.82 N

Please let me know if I'm correct on this.

Sorry for the bad paint skills, and thanks for the reply!
 

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Last edited:
  • #4
Component of g along the inclined plane is g*sinθ, not g*cos θ
 
  • #5
Thanks for the revision!
 

Related to Tension problem on an inclined plane

1. What is tension and how does it affect an inclined plane?

Tension is a force that occurs when an object is pulled or stretched. On an inclined plane, tension is present when an object is being pulled up or down the plane. It can either aid in the motion of the object or resist it, depending on the direction of the applied force.

2. How does the angle of the inclined plane affect tension?

The angle of the inclined plane can greatly affect the amount of tension present. As the angle increases, the component of the weight acting perpendicular to the plane decreases, resulting in a decrease in tension. Conversely, as the angle decreases, the component of the weight acting perpendicular to the plane increases, resulting in an increase in tension.

3. What is the relationship between friction and tension on an inclined plane?

Friction and tension are directly related on an inclined plane. As the angle of the plane increases, the component of the weight acting parallel to the plane increases, resulting in an increase in friction. This also results in a decrease in tension, as the weight is being distributed between friction and tension forces.

4. How does the mass of the object affect tension on an inclined plane?

The mass of the object does not directly affect the tension on an inclined plane. However, as the mass increases, the weight of the object increases, which can result in an increase in the overall tension present on the plane.

5. Can tension ever be greater than the weight of the object on an inclined plane?

No, tension can never be greater than the weight of the object on an inclined plane. Tension is a reactive force and can only be equal to or less than the weight of the object. If the tension were to exceed the weight, the object would start to accelerate in the opposite direction.

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