2 horizontal forces act on a box 2 k.g initially at rest on a smooth table. the first is 3.8N to the north and the 2nd one is 4.7N at an angle of 40 degrees north of east. Find the magnitude and acceleration of the object
thankss...im practicing my physics and i cant get this one done
thankss...im practicing my physics and i cant get this one done
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find the x and y components of each force:
1st force: 3.8N north and 0 east
2nd force: 4.7 sin 40 north and 4.7 cos 40 east
find total x and y components:
total y component = 3.8+ 4.7sin 40 = 6.82N
total x component = 4.7 cos 40 = 3.6N
total force = Sqrt[3.6^2 + 6.82^2] = 7.7N
direction of force is given by tan(theta) = y component/x component = 6.82/3.6
theta = 62.2 deg
the acceleration then is a = F/m = 7.7N/ 2 kg = 3.85 N directed 62.2 deg north of east
1st force: 3.8N north and 0 east
2nd force: 4.7 sin 40 north and 4.7 cos 40 east
find total x and y components:
total y component = 3.8+ 4.7sin 40 = 6.82N
total x component = 4.7 cos 40 = 3.6N
total force = Sqrt[3.6^2 + 6.82^2] = 7.7N
direction of force is given by tan(theta) = y component/x component = 6.82/3.6
theta = 62.2 deg
the acceleration then is a = F/m = 7.7N/ 2 kg = 3.85 N directed 62.2 deg north of east