I don't understand this question 4ia from Jan10 OCR:
Why isn't the answer just 0.4gcos60 and 0.4gsin60 if the particle is at rest? I don't understand why you must calculate the difference
Also, in 7iii in the question below, why does Fr = 7sin30 – 0.42
How come you minus the 0.42, from the force of friction between P and B - I would have thought that that force only acted on P? Because P is the only thing moving, and B is stationary, then wouldn't the force act down the slope (against the direction of motion of P). Sorry this is such a jumble of ideas, but any help is much appreciated?
Thank you in advance!
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Last edited by h8skoooooool; 15-06-2016 at 22:46.
- 15-06-2016 22:40
- 18-06-2016 15:02
For q4 have you taken into the tension in the string due to the second mass?