A box with a mass of 7.00 kg is at rest on a ramp that is at an angle of 20 degr
ID: 1962914 • Letter: A
Question
A box with a mass of 7.00 kg is at rest on a ramp that is at an angle of 20 degrees with the horizontal. The coefficient of static friction between the box and the ramp is 0.800. Use g = 9.80 m/s2. You now want to make the box move by applying a force. To start the box moving, what is the minimum force you need to apply if your force is directed ...(a) parallel to the slope?
.N
(b) perpendicular to the slope?
.N
(c) horizontally?
N
I kow that (a) is The static friction in the x direction minus the box sliding down in the x direction, but I cannot wrap my head around the other two.
Explanation / Answer
Mass m = 7 kg Angle = 20 o The coefficient of static friction between the box and the ramp = 0.800 The minimum force you need to apply if your force is directed parallel to the slope F is equal to sum of static frictional force and weight component along the incline F = Static frictional force + mg sin = mg cos + mg sin = 51.57 N + 23.46N = 75.03 N (b). The minimum force you need to apply if your force is directed perpendicular to the slope F ' then static frictional force is equal to weight component along the incline Static frictional force = mg sin [ mg cos + F ' ] = mg sin mg cos + F ' = mg sin / 64.46 + F ' = 29.325 F ' = -35.135N It is not possible C) Horizontal force F " = ? In this condition F cos = Static frictional force + mg sin = [ mg cos + F sin ] + mg sin 0.9396 F = 51.57+ 0.2052 F + 23.46 0.7344F = 75.03 F = 102.16 N Mass m = 7 kg Angle = 20 o The coefficient of static friction between the box and the ramp = 0.800 The minimum force you need to apply if your force is directed parallel to the slope F is equal to sum of static frictional force and weight component along the incline F = Static frictional force + mg sin = mg cos + mg sin = 51.57 N + 23.46N = 75.03 N (b). The minimum force you need to apply if your force is directed perpendicular to the slope F ' then static frictional force is equal to weight component along the incline Static frictional force = mg sin [ mg cos + F ' ] = mg sin mg cos + F ' = mg sin / 64.46 + F ' = 29.325 F ' = -35.135N It is not possible C) Horizontal force F " = ? In this condition F cos = Static frictional force + mg sin = [ mg cos + F sin ] + mg sin 0.9396 F = 51.57+ 0.2052 F + 23.46 0.7344F = 75.03 F = 102.16 NRelated Questions
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