# Finding Horizontal Force in a Crank at 10N

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• richard9678
In summary, the force in the horizontal direction of a crank can be determined using trigonometric ratios such as sine, cosine, or tangent. However, at 45 degrees, the force does not simply divide into equal horizontal and vertical components. Instead, the horizontal force is 5N, while the vertical force remains at 10N. This highlights the importance of understanding the relationship between angles and forces in this type of problem.

#### richard9678

I'm trying to establish the force in the horizontal direction of a crank when the force that is tangent to the crank is 10N.

Zero degrees is on top of a circle. Clearly, when the crank is vertical which corresponds to it being at zero degrees, the force in the horizontal plane is normal to the crank and is 10N. When the crank is at 90 degrees, the force is vertical and zero in the horizontal.

Now, when the crank is at 45 degrees, the force is acting partly vertical, partly horizontally in equal values, so the the horizontal force should be 5N.

I was hoping to apply either sine, cosine or tangent math to tell me the horizontal force, but neither do. Sin 45 is 0.707, cos 45 is 0.707 and tan 45 is 1. That surprises me that non of the trig. math I'm familiar with has found application in this problem.

richard9678 said:
I was hoping to apply either sine, cosine or tangent math to tell me the horizontal force, but neither do.
You are right that ##\cos{45}=\sqrt{2}/2\approx.707##. How is is that not working?

richard9678 said:
Now, when the crank is at 45 degrees, the force is acting partly vertical, partly horizontally in equal values, so the the horizontal force should be 5N.
This is mistaken. A force of 10N on a diagonal does not break into halves at 5N horizontal and 5N vertical.

richard9678
OK. So, cosθ or sinθ is the correct trig. ratio to use to establish force provided by a crank lever, in the vertical or horizontal as the case may be. Never fully appreciated that at 45 degrees the tangential force does not simply divide into a half of the force at a tangent to the crank. I do now. Thanks.

## 1. What is a crank and why is it important in determining horizontal force?

A crank is a mechanical device that converts rotational motion into linear motion. It is important in determining horizontal force because it is a key component in many machines and mechanisms, such as engines and bicycles, and understanding the force applied to it can help in optimizing its performance.

## 2. How is horizontal force measured in a crank?

Horizontal force in a crank can be measured by using a force sensor or a dynamometer, which can accurately measure the amount of force applied to the crank. Alternatively, it can also be calculated using the principles of torque and mechanical advantage.

## 3. Is there a standard unit for measuring horizontal force in a crank?

Yes, the standard unit for measuring horizontal force in a crank is Newtons (N). This unit is derived from the SI unit for force, which is kilogram-meter per second squared (kg*m/s^2).

## 4. What factors can affect the horizontal force in a crank?

There are several factors that can affect the horizontal force in a crank, such as the angle at which the force is applied, the length of the crank arm, the weight of the object being moved, and any external forces acting on the crank.

## 5. How can the knowledge of horizontal force in a crank be applied in real-world situations?

The knowledge of horizontal force in a crank can be applied in various real-world situations, such as designing and optimizing machines and mechanisms, determining the power and efficiency of engines, and improving the performance of bicycles and other vehicles that use cranks. It can also be used in the field of biomechanics to analyze human movements and improve athletic performance.