What force does Pete exert on the rope?

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To calculate the force Pete exerts on the rope while dragging a crate at a constant speed, one must apply trigonometric principles. Given a horizontal force of 809 N and an angle of 33°, the relationship between the sides of the triangle can be established using sine and cosine functions. Specifically, the sine function relates the opposite side to the hypotenuse, while the cosine function can be used to find the hypotenuse based on the adjacent side. A diagram illustrating the forces acting on the crate can aid in visualizing the problem. Ultimately, this approach will yield the force exerted by Pete on the rope.
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How would I calculate this
A horizontal force of 809 N is needed to drag a crate across a horizontal floor with a constant speed. Pete drags the crate using a rope held at an angle of 33°. What force does Pete exert on the rope?
 
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How about show some working, so we can help you on where you're stuck?

Remember this forum does not do your homework :smile:
 
I was wondering what the formula was?
 
i think u missed some ideas in ur homework. pls clarify it. i think it is work.
 
Use a bit of trigonometry. I'm sure you've learned trigonometry in class.
 
Just think of it as a triangle.

The floor is one length (A) and the rope it the other(H), in this case we make pete the other(O). since we have the angle and we want the length of H we use the following formula. sin33=A/H therefore A/sin33=H just carry on from here.
 
Draw a diagram of th situation. Draw the forces acting on the crate. in this case, since it's moving at constant speed, there is no acceleration.

since the rope is at 33 degrees, it would be a right angled triangle with the adjacent of the 33-degree angle being of magnitude 809

from there, u can use the cosine formula to calculate the hypotenuse of the triangle, which will give u the force exerted by Pete!
 
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