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In each situation below, three pressures are acting on a particle. Given two of the forces and also the fact the fragment is in equilibrium, find the third force.

You are watching: Add a third force that results in a net force of zero.

(mathbfa = quantity-3mathbfi + 2mathbfjN)

(mathbfb = quantity4mathbfi + mathbfjN)

(mathbfa) has magnitude (quantity10N) and direction (45^circ) anticlockwise native the positive (x) axis

(mathbfb= quantity6mathbfi - 3mathbfjN)

In each instance the fragment is in equilibrium, which method the forces are well balanced in all directions and also the net force acting ~ above the fragment is zero. Below we comment on three feasible approaches.

Triangle the forces

Since the particle is in equilibrium we can attract a triangle the forces and also see around what the third force could be.


Using the geometry of the triangle, us can discover the magnitude of the third force with the cosine rule.

<eginalign*mathbfc^2 &= 20^2 + 35^2 - 2 imes 20 imes 35 imes cos60 \mathbf &= 5sqrt37 \&approx quantity30.4Nendalign*>

To describe the 3rd force totally we require the direction that is acting in. Numerous methods could be used, yet we created two best angled triangles that share a side.

<eginalign*5sqrt37cos heta &=20cos 30 \cos heta &= dfrac2sqrt3sqrt37 \ heta &= 55.3^circ ext (3 s.f.)endalign*>

We can draw the pressure on the original diagram.



We can have likewise used Lami’s Theorem, which states that if three forces, acting from a point, space in equilibrium, then the magnitude of each force is proportional come the sine of the angle between the other two forces.

(dfracmathbfsin alpha = dfracmathbfbsin eta = dfracmathbfcsin gamma)

(mathbfa = quantity-3mathbfi + 2mathbfjN)

(mathbfb = quantity4mathbfi + mathbfjN)

It is constantly useful to attract a diagram. The can assist to examine whether answers seem sensible.

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Since the bit is in equilibrium the sum of all the pressures acting ~ above it should be zero, i.e. (mathbfa + mathbfb + mathbfc = 0). Therefore

<-3mathbfi + 2mathbfj + 4mathbfi + mathbfj + mathbfc = 0,>

so the 3rd force is (mathbfc =quantity-mathbfi - 3mathbfjN.)

What are the similarities between this approach and drawing a triangle of pressures as used in the an initial situation?

(mathbfa) has actually magnitude (quantity10N) and also direction (45^circ) anticlockwise from the confident (x) axis

(mathbfb= quantity6mathbfi - 3mathbfjN)

Resolving forces way we break pressures down right into perpendicular components (usually horizontal and vertical). Since it is in equilibrium we know that the sum of the horizontal pressures must be zero, as have to the upright forces. If our third force is (mathbfc = pmathbfi+qmathbfj) then we have the following diagram.


Resolving in the horizontal direction:

Resolving in the vertical direction:

<6 + 10cos45 + p = 0> <10sin45 - 3 + q = 0>

So (mathbfc = left(-6-5sqrt2 ight)mathbfi + left(3 - 5sqrt2 ight)mathbfj.)

What are the similarities and differences between these approaches?

Would every of these approaches work equally well because that the various situations? If not, why not?