Control Lever The 10-lb and 25-lb forces are applied to the control lever of a mechanism. Determine the magnitude and direction of the resultant by using the vector polygon approach. (See Figure 4.8.) Approach We are tasked to find the resultant force on the control lever. We first assume that the

Question

Control Lever

The 10-lb and 25-lb forces are applied to the control lever of a mechanism. Determine the magnitude and direction of the resultant by using the vector polygon approach. (See Figure 4.8.)

Approach
We are tasked to find the resultant force on the control lever. We first assume that the weight of the lever is negligible relative to the applied forces. We then sketch a vector polygon and combine the forces by using the head-to-tail rule. The two given forces, together with the resultant, will form a triangle. We can solve for the unknown length and angle in the triangle by applying the laws of cosines and sines from Appendix B. (See Figure 4.9.)

Accepted Answer

The 25-lb force is sketched first on the diagram, and the 10-lb force is added at the angle of 50° from vertical. The resultant R extends from the tail of the 25-lb force vector (which is labeled as the start point) to the head of the 10-lb force vector (the endpoint). The three vectors form a side-angle-side triangle. Applying the law of cosines, we solve for the unknown side length

[latex]R^2 =(10lb)^2+(25 lb)^2 \longleftarrow [c^2=a^2+b^2-2ab \cos C] [/latex]

[latex]-2 (10 lb)(25 lb) \cos(180° - 40°) [/latex]

from which R = 33.29 lb. The angle [latex]θ[/latex] at which R acts is determined by applying the law of sines

[latex]\frac{\sin(180^\omicron -40^\omicron )}{33.29 lb} =\frac{\sin heta }{10 lb} \longleftarrow [\frac{\sin A}{a}=\frac{\sin B}{b} ] [/latex]

The resultant acts at [latex]θ = 11.13°[/latex] clockwise from the - x-axis.

Discussion
The resultant force is less than the sum of the two forces because the forces are not aligned in the same direction. Also, the direction of the resultant force, at an angle between the two forces, matches the expectation. To double-check the solution, we could break the two forces down into their horizontal and vertical components, as in the technique in Example 4.1. By using the x-y coordinate system, the vector expression for the 10-lb force becomes F = - 7.660 i + 6.428 j lb. See if you can complete the double-check for R and [latex]θ[/latex] by using the vector algebra method.

R = 33.29 lb
θ = 11.13° clockwise from the - x-axis

Question 4.2: Control Lever The 10-lb and 25-lb forces are applied to the ...

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