The Student Room Group

A Level Mechanics

Completely stuck on this question:
Reply 1
Original post by r_g1452
Completely stuck on this question:
Screenshot (285).png
Reply 2
Original post by r_g1452
Screenshot (285).png
You could resolve the forces parallel to the plane
Reply 3
Original post by mqb2766
You could resolve the forces parallel to the plane
I got 15g sin 30 for the lower part of the plane
Reply 4
Original post by r_g1452
I got 15g sin 30 for the lower part of the plane
You mean 15g/2 acts down the plane? So P must provide an equal and opposite force up the plane for equilibrium.

Edit - another similar way is to note you could add the weight and P together tip to tail and the resultant must be perpendicular to the plane which gives a right triangle.
(edited 9 months ago)
Reply 5
Original post by mqb2766
You mean 15g/2 acts down the plane? So P must provide an equal and opposite force up the plane for equilibrium.
image_6209779.JPG So would P sin 30 equal 15g cos 30?
Reply 6
Original post by r_g1452
image_6209779.JPG So would P sin 30 equal 15g cos 30?
I got 15g cos 30 down the plane?
Reply 7
Original post by r_g1452
I got 15g cos 30 down the plane?
You should get Pcos(30) up the plane, so equating that to the resolved weight (down the plane) 15g sin(30) gives P = ....
(edited 9 months ago)
Reply 8
15gsin(30) = 73.5/cos(30) = 84.87N = P ?
Reply 9
Original post by r_g1452
15gsin(30) = 73.5/cos(30) = 84.87N = P ?
agree with the answer, but be careful how you write it down.
Reply 10
Original post by mqb2766
agree with the answer, but be careful how you write it down.
2 significant figures?
Reply 11
Original post by r_g1452
2 significant figures?
No, youve written something like
73 = 84
I know what you mean, but its not correct mathematically.
Reply 12
Original post by mqb2766
No, youve written something like
73 = 84
I know what you mean, but its not correct mathematically.
Right. Thanks 🙏

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