Calculating Force Exerted on a Rope at an Angle

[sharon09]

Okay a horizontal force of 805 N is needed to drag a crate across a horizontal floor with a constant speed. You drag the crate using a rope held at a 32 degree angle.

What force do you exert on the rope?

After I know how to figure this out I can get the rest of the problem myself. I just forgot how to figure this out.

Thanks

 

[Sirus]

I'm assuming you want to continue dragging it at a constant speed. If so, the x-component of the force directed at 32 degrees upwards must be equal to 805 N. Can you figure it out now?

 

[arildno]

Use trigonometry.
Only a component of the tension force (and hence, the force you apply to the rope) is horizontally aligned.

 

[sharon09]

So then is the x-component 805 and the tension force the y-component?

 

[arildno]

So then is the x-component 805 and the tension force the y-component?

No, your tension force can be written as:
$$\vec{T}=T(\cos(32)\vec{i}+\sin(32)\vec{j})$$
The horizontal component is $$T\cos(32)$$
which means that the magnitude of the tension force, T, satisfies:
$$T=\frac{805}{\cos(32)}$$

 

[sharon09]

Thanks so much.
If I want to figure out how much work was done I add the tension force and 805 for the net force right

 

[ponjavic]

I believe the tension is used for creating the 805N force, if this is true then the net force is 805 and the work is displacement * 805 not (805+tension)

 

[Sirus]

Thanks so much.
If I want to figure out how much work was done I add the tension force and 805 for the net force right

Careful! Remember that the 805-Newton force is part of the tension force. As you can see in arildno's post, we are finding the value of the tension force directed at 32 degrees from horizontal necessary for its x-component to be equal to 805 N; that 805 N is the part of the tension force that is directed along the horizontal. Therefore, adding the tension to its x-component to find the net force is incorrect.

I believe the tension is used for creating the 805N force, if this is true then the net force is 805 and the work is displacement * 805 not (805+tension)

The net force acting on the object would actually be the vector sum of the gravitational force and the applied force (tension), assuming no frictional forces are present.

To find the work done, ponjavic you are correct. Since the displacement is in a horizontal direction, we take only the component of the applied force that has done any work, the x-component. Since we have established that that is 805 N, we find that:
$$W=Fd\cos{\theta}=(805 N)d$$
Since the force F in the equation is directed at zero degrees to the horizontal, $\theta=0$, and $\cos{\theta}=1$.

 

[sharon09]

ok thanks so much