Tension in String One of 5 kg Mass at 45 Degree Angle

[UrbanXrisis]

Two strings are attached to a 5 kg mass. String one is at a 45 degree angle and the other is horizontal. What is the tension in string one?

 

[arildno]

From my reading of this, String one is the only one capable of counteracting the weight of the mass (since it alone has a vertical component).

The tension in string one (i.e magnitude of tensile force) should therefore be:
sqrt(2)*mg.

 

[UrbanXrisis]

where did the sqrt(2) come from?

your equation works but how did you get it?

 

[arildno]

Assume that string one goes along the line (-cos(45),sin(45)), wheras string two is along the horizontal (1,0) (The mass itself is position at the origin).

(Clearly, the only necessary requirement is that the tensile force in string one has a component in the positive vertical component, in order to balance the weight of the mass).

 

[KnowledgeIsPower]

T1 = 63.9N
T2 = 49N

I just resolved horizontally and vertically
T1sin45 = 5G
Simplification gives T1 = 63.9N

Calculating T2 is optional, but i got 49N.

 

[KnowledgeIsPower]

To be sure, is the above correct? I'm revising for my mathematics - mechanics exam and that seems to be similar to a few of the questions.

 

[Chen]

Yes that's correct KnowledgeIsPower.

ΣF_x = 0 = T₂ - T₁cos 45^o
ΣF_y = 0 = T₁sin 45^o - mg

From the second equation you can find T₁, and the first will give you T₂.