tow forcess F1=(3.0)in the x direction minus(4.0) in the y direction and F2= (-6.0)in the x direction minus (4.5) in the y direction are applied to a particle. what third force would the the net, or resultant, force on the particle zero?
F1(3,-4)
F2(-6,4.5)
Resultant R is the sum of the component, namely R(-3, 0.5). Resultant is the single force equivalent to the vectorial sum of the other two.
If you are looking to have a third force to keep the particle in equilibrium, you would apply an equal and opposite force to the resultant, namely F=F(3,-0.5).
It is not clear from the question if you are asking for the third force that puts the particle in equilibrium, or the resultant of the two.
A body is at equilibrum under the action of three forces one forces 10N acting dew east and one is 5N In direction 60 north east.what is the magnitude and direction the third
To find the third force that would result in a net force of zero on the particle, we need to add the two given forces together and find the negative of their sum. This is because the negative of the sum of two vectors is equal in magnitude but opposite in direction to the resultant vector that would make the net force zero.
Let's calculate the sum of the two given forces:
F1 = (3.0) in the x direction - (4.0) in the y direction
F2 = (-6.0) in the x direction - (4.5) in the y direction
To add these vectors, we can separately add their x components and their y components:
x components: 3.0 - 6.0 = -3.0
y components: -4.0 - 4.5 = -8.5
Now, we have the resultant vector as (x: -3.0, y: -8.5). To find the third force that cancels out this resultant vector, we need to take the negative of each component:
Third force: (-x: 3.0, -y: 8.5)
Therefore, the third force required to make the net force on the particle zero is 3.0 in the positive x direction and 8.5 in the positive y direction.