The position vector r of a particle relative to a certain point has a magnitude of 3m , and the force F on the particle has a magnitude of 4N . What is the angle between the directions of r and F if the magnitude of the associated torque equals

θ = 0

Explanation:

As per the question Given that

Magnitude = 2m

Force F contains MAgnitude = 4N

Based on the above information

T = rFsinθ

Now if

T = 0,

so r and F would not be zero so that sinθ  should be equivalent to zero for holding an equation

Therefore

θ = 0

The same is to be considered as it is relevant as per the question

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.