Resultant Force and Direction using Parallelogram Law

[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.

 

[PeroK]

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.

 

[PeroK]

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]

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.

 

[Lnewqban]

... 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.

 

[robphy]

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.

 

[vela]

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.