A Boy Pulls His Toy On A Smooth Horizontal Surface With A Rope Inclined At 60 Degree To The Horizontal.If The Effective Force Pulling The Toy Along The Horizontal Surface Is 5N,calculate The Tension In The Rope:
Divide W which is 5 by COS 60 it gives 10 that is 10N
well T*cos60° = 5, so ...
why all the capitals???
10 N
10N
Well, well, well, it seems we have a boy and his toy in a bit of a pickle, huh? Alright, let's see if Clown Bot can help!
Now, if I understand correctly, the rope is inclined at 60 degrees to the horizontal and the boy is pulling the toy with a force of 5 N. We need to calculate the tension in the rope.
Alright, here comes the magic (or math, if you insist). Since the rope is inclined, we can break down the force into two components: one along the horizontal surface and the other along the inclined rope.
The force pulling the toy along the horizontal surface is the effective force of 5 N. But we need to know the force along the inclined rope. Luckily, we have some trigonometry up our sleeve!
Using some cosine magic, we can find that the force along the inclined rope is given by:
Force along the inclined rope = Effective force / cos(60 degrees)
Now, let's plug in the numbers and perform our circus act!
Force along the inclined rope = 5 N / cos(60 degrees)
Using our math tricks, we find that the force along the inclined rope is approximately 10 N. Ta-da!
So, the tension in the rope is around 10 Newtons. Just remember, kids, it's all fun and games until someone asks Clown Bot for math help!
To calculate the tension in the rope, you can use trigonometry.
First, we need to break down the forces acting on the toy. We have the effective force pulling the toy along the horizontal surface, which is 5N. We also have the tension in the rope, which we need to find.
Since the rope is inclined at 60 degrees to the horizontal, we can use the component of the tension force in the horizontal direction to balance the effective force of 5N.
To find this component, we will use the trigonometric function cosine, which relates the adjacent side and the hypotenuse of a right triangle. In this case, the tension force in the rope is the hypotenuse, and the horizontal component is the adjacent side.
So, we can use the equation:
cos(60 degrees) = Adjacent side / Hypotenuse
Simplifying that equation:
0.5 = Adjacent side / Tension force
Rearranging the equation to solve for the tension force:
Tension force = Adjacent side / 0.5
Since the adjacent side represents the horizontal component of the tension force and is equal to the effective force pulling the toy (5N), we can substitute this value into the equation:
Tension force = 5N / 0.5
Calculating:
Tension force = 10N
Therefore, the tension in the rope is 10N.