Struggling with this PreCalc question about finding torque with cross products.

Fnun
timer Asked: Apr 6th, 2018

Question Description

Question:

A mechanic applies a force of 42 Newtons straight down to a ratchet that is 0.59 meter long. What is the magnitude of the torque when the handle makes a 38° angle above the horizontal?

There is an example problem, but I'm not sure how to do it either, I'll attach it for reference.

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Example 4: Torque Using Cross Product 40° 25N AUTO REPAIR Robert uses a lug wrench to tighten a lug nut. The wrench he uses is 50 centimeters or 0.5 meter long. Find the magnitude and direction of the torque about the lug nut if he applies a force of 25 newtons straight down to the end of the 0.5 m handle when it is 40° below the positive x-axis as shown. Step 1 Graph each vector in standard position (Figure 8.5.1). Step 2 Determine the component form of each vector. The component form of the vector representing the directed distance from the axis of rotation to the end of the handle can be found using the triangle in Figure 8.5.2 and trigonometry. Vector r is therefore (0.5 cos 40°, 0, -0.5 sin 40') or about (0.38, 0, -0.32). The vector representing the force applied to the end of the handle is 25 newtons straight down, so F = (0,0,–25). 0.5 cos 40 40 0.5 sin 40 40" | | = 0,5 m Figure 8.5.1 Figure 8.5.2 Step 3 Use the cross product of these vectors to find the vector representing the torque about the lug nut. T=rXF Torque Cross Product Formula 0.38 0 -0.32 Cross product of rand F 0 0 -25 10 -0.32 -10.38 -33|3 +10.38 OK Determinant of a 3 x 3 matrix -25 = 0i - (-9.5)j + OK Determinants of 2 x 2 matrices = (0,9.5, 0) Component form Step 4 Find the magnitude and direction of the torque vector. The component form of the torque vector (0,9.5, 0) tells us that the magnitude of the vector is about 9.5 newton-meters parallel to the positive y-axis as shown. F 40
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