A boy pulls a wagon with a force of 20N on a handle of .50 meters long. The end he holds is .25 meters higher than the end attached to the wagon. What is the magnitude of the horizontal component that acts to pull the wagon forward?
Divide .25m/.50m for a total of .50
This is sort of like cosθ=adj/hyp=.25/.50=.600
Cos^-1 for .50 and your angle is 60°.
After that do 20N·sin 60°
This is hyp· sin θ
Your answer being 17N
Will someone answer this please. Thank you.
No haha
Why did the wagon go to therapy? Because it had an unresolved horizontal component issue! But don't worry, I'm here to help. Let's calculate that horizontal component for you.
To find the magnitude of the horizontal component, we need to look at the angle between the handle and the horizontal plane. Since the difference in height between the two ends of the handle is given, we can use trigonometry to find the angle.
The difference in height is 0.25 meters, and the length of the handle is 0.50 meters. So, the angle (θ) is given by the inverse tangent (arctan) of the opposite side divided by the adjacent side.
θ = arctan(0.25/0.50)
Now that we have the angle, we can find the horizontal component of the force. The horizontal component can be found by multiplying the force applied (20N) by the cosine of the angle.
Horizontal component = 20N * cos(θ)
Calculate the value of cos(θ) using the angle we found earlier, and you'll have your answer!
To find the magnitude of the horizontal component that acts to pull the wagon forward, we can use the concept of torque. Torque is defined as the product of the force applied and the perpendicular distance from the point of rotation.
In this case, the force applied by the boy is 20N, and the handle of the wagon is 0.50 meters long. We can assume that the point of rotation is the end attached to the wagon.
First, we need to find the perpendicular distance between the point of rotation and the force. Since the end held by the boy is 0.25 meters higher, the perpendicular distance is given by the Pythagorean theorem:
Perpendicular distance = √(length of handle² - height difference²)
= √(0.50² - 0.25²)
= √(0.25 - 0.0625)
= √0.1875
= 0.433 meters
Now, we can find the magnitude of the horizontal component using the equation:
Torque = Force x Perpendicular distance
The torque applied by the boy is given by:
Torque = 20N x 0.433 meters
= 8.66Nm
The magnitude of the horizontal component, which acts to pull the wagon forward, is equal to the torque applied. Therefore, the magnitude of the horizontal component is 8.66N.