Question

Consider a force \(F{\bf{ = 80}}\;{\bf{N}}\) applied to a beam as shown in Fig. 8–37. The length of the beam is \(l{\bf{ = 5}}{\bf{.0}}\;{\bf{m}}\) and \(\theta {\bf{ = 3}}{{\bf{7}}^{\bf{o}}}\), so that \(x{\bf{ = 3}}{\bf{.0}}\;{\bf{m}}\) and \(y{\bf{ = 4}}{\bf{.0}}\;{\bf{m}}\). Of the following expressions, which ones give the correct torque produced by the force around point P?

(a) 80 N.

(b) (80 N)(5.0 m).

(c) (80 N)(5.0 m)(sin 37°).

(d) (80 N)(4.0 m).

(e) (80 N)(3.0 m).

(f) (48 N)(5.0 m).

(g) (48 N)(4.0 m)(sin 37°).

FIGURE 8-37MisConceptual Question 5.

Short answer

The correct options are (c), (e), and (f).

Step-by-step solution

Magnitude of torque

The magnitude of torque is obtained by the multiplication of force and its perpendicular distance from the pivot point.

Calculation of torque

The perpendicular distance of the force \(F = 80\;{\rm{N}}\) from the pivot point is \(x = 3.0\;{\rm{m}}\).

Therefore, the torque is:

\(\begin{aligned}\tau &= Fx\\ &= \left( {80\;{\rm{N}}} \right) \cdot \left( {3.0\;{\rm{m}}} \right)\end{aligned}\)

Therefore, option (e) is correct.

Now, from the figure, you get:

\(\begin{aligned}\frac{x}{l} &= \sin \theta \\x &= l\sin \theta \end{aligned}\)

Then, the torque is:

\(\begin{aligned}\tau &= Fx\\ &= Fl\sin \theta \\ &= \left( {80\;{\rm{N}}} \right) \cdot \left( {5.0\;{\rm{m}}} \right)\sin {37^ \circ }\end{aligned}\) … (i)

Therefore, option (c) is also correct.

Now, from equation (i), you get:

\(\begin{aligned}\tau &= \left( {80\;{\rm{N}}} \right) \cdot \left( {5.0\;{\rm{m}}} \right)\sin {37^ \circ }\\ \approx \left( {48\;{\rm{N}}} \right) \cdot \left( {5.0\;{\rm{m}}} \right)\end{aligned}\)

Therefore, option (f) is also correct.