Finding the force balancing a weight

[jsmith613]

Homework Statement

http://www.edexcel.com/migrationdocuments/GCE New GCE/GCE-Physics-6PH01.pdf
Q22)

Homework Equations

-

The Attempt at a Solution

Look at Q22)

I could do a and b but had issues with c)

for ci) would I be correct in thinking:
- The dotted line is "tension"
- the line straight down from the dot is "weight"
- the dark, diagonal line is contact force
?

for cii)
I would have said that weight = sin(40) * h
so 640 / sin(40) = h = 995.66

BUT
The markscheme says that the force balancing weight is 640 * sin(40) = 411.4


Can someone help me understand all this?

 

[cepheid]

for ci) would I be correct in thinking:
- The dotted line is "tension"
- the line straight down from the dot is "weight"
- the dark, diagonal line is contact force
?

I agree with all of this. Note that there is no "straight line down from the dot" in the diagram, but I agree that you need to draw one in that represents the climber's weight.

I have no idea what the fourth force should be either (as far as I can tell, those are the three that are present).

for cii)
I would have said that weight = sin(40) * h
so 640 / sin(40) = h = 995.66

BUT
The markscheme says that the force balancing weight is 640 * sin(40) = 411.4


Can someone help me understand all this?

What is "h" supposed to be? In any case, here is my approach to solving the problem. For your answer to part (a), you should have said that "equilibrium" means that the net force on the climber is zero. In other words, the vector sum of all forces acting on the climber is zero. This means that the sum of the component forces in the x-direction is zero, and the sum of the component forces acting in the y-direction is also separately zero:

∑F_x = 0
∑F_y = 0

This gives you two equations, and since you have two unknowns (the magnitude of the tension and the magnitude of the normal force) you can arrive at a solution. Just start by resolving the normal force and the tension into x and y components.

 

[jsmith613]

sorry, h was supposed to be t which is tension.

If we imagined a right-angled triangle then the angle between the base and hypotenuse is 40, the height is 640 and the diagonal is tension, t.

As we have opp and hypotenuse, we use sin.
As we want hypotenuse we say opp/sin(angle) so 640/sin(40) but this is wrong and I can't see why?

 

[cepheid]

I'm assuming that you think the height of the triangle should be 640 because it must equal the weight? I would say that there is the flaw in your reasoning. It should be clear from the free-body diagram that the tension is not the only force that has an upward vertical component, and therefore it is not the only force that is available to balance the weight. That is why it is very important to start with

∑F_y = 0

and take any inventory of everything that acts vertically.

 

[jsmith613]

But the only force I am intereseted in is tension because that's what they are after.

Anyhow, the two vertical forces would be of tension and contac
SO
= 640 / 2 * sin(40)
= 497.8 which is way off

So what's going wrong?

 

[jsmith613]

Problem to last question solved:

New question:
http://www.ocr.org.uk/download/pp_10_jan/ocr_49353_pp_10_jan_gce_g481_ml.pdf

Q3bi)
On looking at the mark-scheme it is suggested that the total tension =

total mass of all components = 1480
1480(9.91 + 1.8)
OR
weight of lift + net force due to accelaration

BUT
I would have thought we would say 1480(9.81 - 1.8), i.e take away net force as this is in the oposite direction to weight so the resultant force would be smaller?

 

[RCB]

yeh ,surely they should subtract from each other not add??