Weight of an object in static equilibrium

[Zack K]

Homework Statement

An object is suspended on a frictionless slope inclined plane by a rope parallel to the incline. If the angle of the incline is 25° and the tension on the rope is 5000 N, what is the weight of the object

Homework Equations

∑F=0

The Attempt at a Solution

So in my eyes, you would want to find the vertical component of the tension of the rope, which is counteracted by the weight of the object. Getting the y-component you would do 5000sin25 to get 2113 N. So shouldn't the weight of the be 2113 N, since ∑F_y=0?

 

[Chestermiller]

Homework Statement

An object is suspended on a frictionless slope inclined plane by a rope parallel to the incline. If the angle of the incline is 25° and the tension on the rope is 5000 N, what is the weight of the object

Homework Equations

∑F=0

The Attempt at a Solution

So in my eyes, you would want to find the vertical component of the tension of the rope, which is counteracted by the weight of the object. Getting the y-component you would do 5000sin50 to get 2113 N. So shouldn't the weight of the be 2113 N, since ∑F_y=0?

What about the vertical force that the incline exerts on the object? You need to determine the mass of the object.

 

[Zack K]

What about the vertical force that the incline exerts on the object? You need to determine the mass of the object.

I'm not entirely sure on how to find mass since to my knowledge to find mass you would need the normal force/weight.

 

[Chestermiller]

You need to perform a force balance on the object in the direction parallel to the slope. Have you drawn a free body diagram?

 

[Let'sthink]

Every concept in Physics has two definitions. one theoretical and other operational or experimental. The theoretical gives its relation with other concepts usually it can be expressed by a mathematical formula. Operational is a process by which to measure it. So weight is what you would get if you hold the objects by a vertical spring balance. Here you are holding it on an incline. So think about what Chestermiller is telling you.

 

[Let'sthink]

Homework Statement

An object is suspended on a frictionless slope inclined plane by a rope parallel to the incline. If the angle of the incline is 25° and the tension on the rope is 5000 N, what is the weight of the object

Homework Equations

∑F=0

The Attempt at a Solution

So in my eyes, you would want to find the vertical component of the tension of the rope, which is counteracted by the weight of the object. Getting the y-component you would do 5000sin50 to get 2113 N. So shouldn't the weight of the be 2113 N, since ∑F_y=0?

 

[Let'sthink]

where from you have got 50? Better draw free body diagram correctly. Note the forces acting on the suspended object. One is tension in the rope. Forces are not visible they have to be given, felt or imagined.

 

[Zack K]

You need to perform a force balance on the object in the direction parallel to the slope. Have you drawn a free body diagram?

I've just drawn it. So there would be 3 forces acting on the object right? Tension, gravity and the force the incline exerts on the object, which would be 5000cos(25). So now would I find the y-components of the tension and the objects normal force and set it equal to the force of gravity?

 

[Chestermiller]

I've just drawn it. So there would be 3 forces acting on the object right? Tension, gravity and the force the incline exerts on the object, which would be 5000cos(25). So now would I find the y-components of the tension and the objects normal force and set it equal to the force of gravity?

The force the incline exerts on the object is not 5000 cos(25). The force the incline exerts on the object is $mg \cos(25) = W \cos(25)$. That follows for the force balance normal to the incline. Now, please write your force balance equation for the direction parallel to the incline, making use of the component of the gravitational force parallel to the incline.

 

[Zack K]

The force the incline exerts on the object is not 5000 cos(25). The force the incline exerts on the object is $mg \cos(25) = W \cos(25)$. That follows for the force balance normal to the incline. Now, please write your force balance equation for the direction parallel to the incline, making use of the component of the gravitational force parallel to the incline.

Oh jeez sorry that was another typo I meant mgcos(25). So my balance equation would be, 5000sin(25) + mgcos(25)sin(65)(for the y-component)= mg. The left hand side is all the upwards forces and the right hand side is all the downward forces. Then you solve for mg.

 

[Let'sthink]

You need to decide are you working with x y resolution of forces or parallel to incline and perpendicular to incline. Both will give you correct result but the latter is more straightforward and less confusing. You have mixed up he two but it is ok. Still your equation is right now just put sin 65 = cos 25 and simplify you will get what you would get directly also. I appreciate your understanding of Chestmiller and following it correctly. But think coolly and sort out muddled up equations and issues in this problem. Think about them aloud and write here. I would like to help you in clarifying certain issues. Once you sort them out mentally for this problem it will help you in other problems also.

 

[Chestermiller]

I agree with @Let'sthink . Even though you have ended up getting the right answer by doing a force balance on the object in the vertical direction, it would have been so much easier for you to obtain the same final result by doing a force balance on the object in the direction parallel to the slope. Please write that down for us so that we can be sure you have benefited from what Let'sthink has said.

 

[Let'sthink]

I agree with @Let'sthink . Even though you have ended up getting the right answer by doing a force balance on the object in the vertical direction, it would have been so much easier for you to obtain the same final result by doing a force balance on the object in the direction parallel to the slope. Please write that down for us so that we can be sure you have benefited from what Let'sthink has said.

Thank you Chestermiller. I wish to make one more point here. It is true there are three force acting on the object which will be also shown in the free body diagram. One must understand out of these three the only external force which is fixed and which causes the other two adjustable forces is the pull by Earth on the object. So if that is the causing the other two then let us resolve it in two components one along which motion is possible and other along which motion is not possible.