3.3.4 Newton's third law and equilibrium

Question 3

An anchoring force $P\mathbf{P}$ of magnitude 120 N is applied to a structural joint. The force acts in the third quadrant relative to the standard unit vectors $i\mathbf{i}$ (pointing right) and $j\mathbf{j}$ (pointing upwards). The direction of the force makes an angle of $25^\circ\,$ with the negative $y$ -axis.

The force can be expressed as a vector $N\begin{bmatrix} P_x \\ P_y \end{bmatrix} \text{ N}$ .

Find the correct expression for $P_y$ .

$Py=120cos⁡25∘P_y = 120 \cos 25^\circ$

$Py=−120sin⁡25∘P_y = -120 \sin 25^\circ$

$Py=−120cos⁡25∘P_y = -120 \cos 25^\circ$

$Py=120sin⁡25∘P_y = 120 \sin 25^\circ$

[3]

3.3.4 Newton's third law and equilibrium

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Question 3

An anchoring force $P\mathbf{P}$ of magnitude 120 N is applied to a structural joint. The force acts in the third quadrant relative to the standard unit vectors $i\mathbf{i}$ (pointing right) and $j\mathbf{j}$ (pointing upwards). The direction of the force makes an angle of $25^\circ\,$ with the negative $y$ -axis.

The force can be expressed as a vector $N\begin{bmatrix} P_x \\ P_y \end{bmatrix} \text{ N}$ .

Find the correct expression for $P_y$ .

$Py=120cos⁡25∘P_y = 120 \cos 25^\circ$

$Py=−120sin⁡25∘P_y = -120 \sin 25^\circ$

$Py=−120cos⁡25∘P_y = -120 \cos 25^\circ$

$Py=120sin⁡25∘P_y = 120 \sin 25^\circ$

[3]

3.3.4 Newton's third law and equilibrium

Question 3

3.3.4 Newton's third law and equilibrium Questions

3.3.4 Newton's third law and equilibrium

Question 3

3.3.4 Newton's third law and equilibrium Questions