Physics Help Forum Need quick help with tension problem

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 Oct 25th 2012, 10:36 PM #1 Junior Member   Join Date: Oct 2012 Posts: 7 Need quick help with tension problem As far as explanitions. I'm in physics 1 and this is my 4th chapter. I've learned basic kinematics, basic friction and basic tension, basically, I'm a beginner, so any explanitions with basic stuff will be helpful A 205 kg log is pulled up a 30 degree ramp by a rope parallel to the ramp. The coefficient of kinetic friction is .9 and the log has an acceleration of .8 m/s2. Find the rope tension Attempt: I honestly have no idea where to start on this at all. My teacher doesn't do a very good job of going over everything, so I have no examples of calculating the tension of a rope pulling an object, nor do I have one of an inclined plane. Any and all help you could give me would be a huge help to me. I have one more, but its a little hard to explain. I have a triangle, the lowest part of the hypontunse is on the left, and the highest part is on the right. The second longest side(horizontal) connects to the lowest part of the hypontunse to form a 15 degree angle. And then there is a third(completly vertical) side connecting all three. I have a 15 kg box on the ramp being pushed with an acceleration of .5 m/s2. The coefficient of kinetic friction is .43, what is the force. Attempt: NO idea where to start. Looking at all of my problems, all we did was solve for acceleration. This is the first I'm seeing of a problem where we solve for force on an inclined plane
 Oct 26th 2012, 06:24 AM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,352 For the first problem, whenever you have something on a ramp it's almost always a good idea to split all the forces, accelerations, velocities, etc into components acting along the direction of the ramp and those acting perpendicuar to the ramp. If the ramp is at angle theta to the horizontal then the log's weight, which is a force acting straight down, can be resolved into a component acting perpendicular to the ramp (or "normal" to the ramp, as we like to say) - which is equal to its weight times cosine theta - and a component acting parallel to the ramp - which is the weight times sine theta. Do you see why that is? It's very important to get this bit correct. Now consider the equation: sum of forces = ma. In the direction parallel to the ramp you have the following forces: tension on the rope acting in the + direction (i.e. pulling up the ramp), and both friction and the component of the log's weight that is parallel to the ramp that resists the tension hence pulling in the negative direction. The friction force is equal to the coefficient of friction (u) times the normal component of the log's weight (W), or F_friction= uW cos(theta). The component of weight due to gravity is F_grav = W sin(theta). So now you can write out all the elements of F: F= T-W sin(theta) - uW cos(theta) Set this to ma and solve for the T. Post back with what you get and we'll review it for you. The second problem is very similar. Again, consider forces acting along the direction of the 15 degree ramp - you have the pushing force, you have the component of the box's weight resisting that force, and the friction force also resisting. As above, apply F=ma and solve for T. Last edited by ChipB; Oct 26th 2012 at 06:27 AM.
 Oct 26th 2012, 11:00 AM #3 Physics Team     Join Date: Jun 2010 Location: Mauritius Posts: 609 Maybe that picture will help? Along the plane: Ff is the force due to friction, mg sinθ is the component of weight, T is the tension. Perpendicular to the plane: mg cosθ is the component of weight, R is the normal reaction force. Perpendicular to the plane, we don't have any net force (or the log would sink into the ramp, or jump over it), hence R = mg cosθ Along the plane, we have the tension T being the largest of all the forces. T - mg sinθ - Ff = ma You will have to realise that θ = 30 degrees And that Ff = μR = μ(mg cosθ) [μ - coefficient of friction] So that: T - mg sinθ - μ(mg cosθ) = ma For the second problem, you will have to be careful about the direction of the force; whether it is along the plane or at the horizontal, or at any other angle to the horizontal. Since you didn't mention anything about that, I would assume it's along the plane, and the easier way of the different possibilities. It would be like the mirror image of the picture I posted above, with only the mass and angle changing, and a force F instead of the tension T. __________________ Jerry (Got my results!) It is easier to protect your feet with slippers than to cover the earth with carpet. No one can go back and change a bad beginning; but anyone can start now and create a successful ending. If a problem can be solved, no need to worry about it. If it cannot be solved what is the use of worrying?

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# how to find the tension of a rope on a inclined plane

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