A block of mass 5 kg rests on a rough plane inclined at 35∘ 35^\circ\,35∘ to the horizontal, as shown in the diagram.
The unit vectors i\mathbf{i}i and j\mathbf{j}j are defined such that:
- i\mathbf{i}i is perpendicular to the inclined plane, pointing upwards and away from the surface (parallel to the normal reaction force R\mathbf{R}R).
- j\mathbf{j}j is parallel to the inclined plane, pointing up the slope (parallel to the friction force F\mathbf{F}F).
Taking g g\,g as the acceleration due to gravity, which of the following is the correct vector expression for the weight force W\mathbf{W}W of the block?
W=−5gsin35∘i−5gcos35∘j\mathbf{W} = -5g \sin 35^\circ \mathbf{i} - 5g \cos 35^\circ \mathbf{j}W=−5gsin35∘i−5gcos35∘j
W=−5gcos35∘i−5gsin35∘j\mathbf{W} = -5g \cos 35^\circ \mathbf{i} - 5g \sin 35^\circ \mathbf{j}W=−5gcos35∘i−5gsin35∘j
W=−5gcos35∘i+5gsin35∘j\mathbf{W} = -5g \cos 35^\circ \mathbf{i} + 5g \sin 35^\circ \mathbf{j}W=−5gcos35∘i+5gsin35∘j
W=5gsin35∘i−5gcos35∘j\mathbf{W} = 5g \sin 35^\circ \mathbf{i} - 5g \cos 35^\circ \mathbf{j}W=5gsin35∘i−5gcos35∘j