# Resultant Force and Direction using Parallelogram Law

• Engineering
• Nova_Chr0n0
Nova_Chr0n0
Homework Statement
2-79. Use the parallelogram law to determine the magnitude of the resultant force acting on the pin. Specify the resultant's direction, measured from the x-axis.
Relevant Equations
Cosine Law
Sine Law

Picture above is the complete question. I want to ask about the problem where I would use the parallelogram method. Here is my FBD:

I start off by computing the angle alpha:
α + α + 105 +105 = 360
α = 75 degree

After that, I now use cosine law to solve for the resultant force:
R=[(80)^2+(50)^2-2(80)(50)cos(75)]^(1/2)
R = 82. 640 lb

For the resultant angle with respect to the x-axis, I tried computing for angle beta first:
[sin(β)/80] = [sin(75)/82.640]
β = 69.239

To get theta, I now subtract angles and got:
θ = 90-30-69.239
θ = -9.239 degree

Here where my question starts, my initial thought is that the negative value is acceptable since the angle I'm finding is below the horizontal. But when I did the component method instead of the parallelogram method, I got the position of theta to be wrong.

USING COMPONENT METHOD:
Rx = 80sin(45)+50sin(30) = +81.569 lb
Ry = 80cos(45)-50cos(30)= +13.267 lb

Based on this value, The x component of the resultant force is (+) and the y component is also (+). So the resultant force should be above the x-axis. But in the drawing of my parallelogram, it is below the +x-axis. Is my drawing of the parallelogram inaccurate? or am I missing an important information/knowledge here? If I continue the solution for the component method, I still got 9.239 degree as my answer. But the position the theta in my drawing worries me.

Your diagram is not quite right, because you've shown the resultant force is below the x-axis. You got ##\beta = 69## degrees, which gives a resultant angle of about ##9## degrees above the x-axis.

Personally, I would use vector components to calculate the force before I drew the diagram. For the good reason that, until you've done the calculation, you don't know where to draw the resultant force.

Nova_Chr0n0
PS I could look up what the "parallelogram" law is, but it's not something I use. The cosine rule is useful; as is the sine rule.

Nova_Chr0n0
PeroK said:
Your diagram is not quite right, because you've shown the resultant force is below the x-axis. You got ##\beta = 69## degrees, which gives a resultant angle of about ##9## degrees above the x-axis.

Personally, I would use vector components to calculate the force before I drew the diagram. For the good reason that, until you've done the calculation, you don't know where to draw the resultant force.
Thanks! Upon reviewing it, my diagram is actually wrong. I have now corrected it and the resultant drawing using parallelgoram method now lands on the upper part of the x-axis. I'm also not a fan of using parallelogram espsecially when resultants are involved. But well, I got to follow the instruction given.

Nova_Chr0n0 said:
... I have now corrected it and the resultant drawing using parallelogram method now lands on the upper part of the x-axis.
Note that if the magnitude of the 50 lbf was smaller, the angle of the resultant force respect to the x-axis would be greater.

Nova_Chr0n0
One reason to use the parallelogram law (as in 2-79) [tails together, draw parallels through the tips of the vectors, then construct the diagonal from the tails to the opposite corner] is to construct the resultant with geometrical tools (a ruler), which can be approximated with a sketch. (No calculator is needed to get the [approximate] direction... one can develop some geometric intuition for the resultant.)

Use the side-lengths and angles with either components (as in 2-80) or the sine and cosines laws for a more precise calculation.

Nova_Chr0n0, Lnewqban and vela
Nova_Chr0n0 said:
Is my drawing of the parallelogram inaccurate? or am I missing an important information/knowledge here?
You've hopefully already realized this, but I'll explicitly state it for the benefit of other students who run across this thread since it's a common mistake in my experience. You sketched the two vectors with approximately the same length, but from the information given, you should have made the bottom vector about 2/3 the length of the other vector. When you're making a sketch and using the parallelogram rule, it's critical that you draw the vectors to scale.

Nova_Chr0n0, Tom.G, Lnewqban and 1 other person

## What is the Parallelogram Law of Forces?

The Parallelogram Law of Forces states that if two vectors acting simultaneously at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram, then their resultant vector is represented by the diagonal of the parallelogram that passes through the same point.

## How do you determine the magnitude of the resultant force using the Parallelogram Law?

To determine the magnitude of the resultant force, you can use the formula: R = √(P² + Q² + 2PQcosθ), where P and Q are the magnitudes of the two forces, and θ is the angle between them. This formula is derived from the law of cosines applied to the parallelogram.

## How do you find the direction of the resultant force using the Parallelogram Law?

The direction of the resultant force can be found using the formula: tan(α) = (Qsinθ) / (P + Qcosθ), where α is the angle between the resultant force and one of the original forces (P). This formula is derived from trigonometric relationships within the parallelogram.

## What is the significance of the angle between the two forces in the Parallelogram Law?

The angle between the two forces significantly affects both the magnitude and direction of the resultant force. When the angle is 0 degrees (forces are in the same direction), the resultant force is at its maximum. When the angle is 180 degrees (forces are in opposite directions), the resultant force is at its minimum. For any other angle, the resultant force will be somewhere between these extremes.

## Can the Parallelogram Law be used for more than two forces?

The Parallelogram Law specifically applies to the combination of two forces. For more than two forces, you can use vector addition methods such as the polygon method or the component method. These methods involve breaking down the forces into components, adding them up, and then finding the resultant vector.

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