Question

A 15kg object lies on a frictionless (icy) horizontal table. A 30N force pushes on it to the north. At the same time, a steady wind pushes on it with a force of 9N to the west. (Let the y-axis be north and x-axis be east, use traditional x-y coordinate system).

a) What is the total force on the object and in what direction is the force?
b) If the object is held still and then let go, how far from the starting point will it be in 2,5 seconds?
c) If the table is not frictionless (not icy) but has the coefficient of friction og u=0,120 to the object, what is the total force on the object and in what direction?
d) It's possible to make the object stay in place by adding a second object on top of it. What is the minimum mass that needs to be added to make the object stable? u=0,120

Answer 1

Sam Given, mass of object m= 15kg. two forces are acting on the object in north and west directions. y (North) 1. BON INA Iskg (E951) we want to calculate total force so, to tel force is equal to vector sum of so let 3 = 3o individual force. P = gi) = -gi Resultant force f = A + F Magnitude IFI = √(30) ²+1992 TE= 31.32 New tome direction. ex tamo - 30 50,[9= 43.30 alirection of resulfonst force is 106.7 angle from 4x axis. 16 let the object be at origin (0,0) at time tao. then, the ase forces work as given in question:

. We know that force in north direction accelerate the object in north direction and force in west direction will accelerate to the object in west direction So, object af tey 2.5 see, letts, yo coordin toate of object- • y = lyt + 2 ayt? given, "yão, ay = 2 2 2 2 3 2 2 2 27/02 so, y = alt) + 1 (2) (2.594 yä 625 m I now, a-coordinate x= 4 + + b bet2 given, 45x20, = 706 ms2 -0.6 mshoz so, x = 0.16) + 46016) (2.532 Tas -1.875 ml so, object will be at (-1.875, 6:25) after 2.5 seconds. Now; cofficient of friction M= 0.120 So, friction force will adoo also act on object in just opposite to wet force (due to i and 2).

force Net 17)=3132N object We know that friction force oppose relative velocity between two surface. friction force 17)=18N magnitude of friction force = un (10) = 150 Newton So, f = (0.20 2(150) = 18 New son wet so, Net force after adding foiction force will be equal to 7- = (3182)-(18) = 13:22 (f) - Parker = 13.82 Newtons direction is given in dingsam. - 18.32 No 13.82 Newtons en in diagram. - NO 13 IP in 13-32 Newtons. her alirection

Now we want to make object stable so we need to increase value of friction force by adding mass. , letts we add a mass m them (M +18)g (0:120) = 81.32 M +15=26.10 M = 11.1kg so, Added mass should be equal to 111 kg.