What are the horizontal and vertical components of a

300 n force that is applied along a rope at 60° to the

horizontal and used to drag an object across a yard?

! can u explain the solve? idk why my answer is always reverse, like the answer for horizontal, not for vertical. the answer for vertical, not for horizontal. so, i am confusing. pls !

if u got the answer,

In the figure attached we can see in a better way this situation. If we are told the rope is at 60 degress to the horizontal, we can place this rope in the origin of the X-Y coordinate system.

If we observe in detail, we have a right triangle where the opposite side to the angle is the vertical component and the adjacent side is the horizontal component . So, we can use trigonometric functions to find the force (the hypotenuse ) applied to the rope.

For :

We can use the trigonometric function cosine, which is defined as:

This means

For :

We can use the trigonometric function sine, which is defined as:

This means

Finally the horizontal and vertical components are 150 N and 259.8 N

the force on the thing and the distance the force moves the thing

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.