A horizontal force P is exerted on a 20kg box to slide it up a 30 degree incline. The friction force regarding motion is 80N. How large must P be if the acceleration of the moving box is to be (a) zero and (b) 0.75 m/s^2?
The answer is (a) 206N and (b) 223N, but I wanna know how to get those answers.....
The answer is (a) 206N and (b) 223N, but I wanna know how to get those answers.....
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P is exerted on a 20kg box to slide it up a 30 degree incline.
a)
Frictional force acts down the plane.
For equilibrium P cos θ = mg sin θ + 80
P cos 30 = 20*9.8* sin 30 + 80
P = 206 N
b)
When there is an acceleration of 0.75 m/s^2
An additional force of ma/ cos 30 has to be applied = 20* 0.75 /cos 30 = 17 N
Total force 206 + 17 = 223 N
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a)
Frictional force acts down the plane.
For equilibrium P cos θ = mg sin θ + 80
P cos 30 = 20*9.8* sin 30 + 80
P = 206 N
b)
When there is an acceleration of 0.75 m/s^2
An additional force of ma/ cos 30 has to be applied = 20* 0.75 /cos 30 = 17 N
Total force 206 + 17 = 223 N
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I'm not too good with inclined planes, but I'll try my best with how I interpret it in terms of a flat surface.
Σ=Sum of
f = friction
P=Force of pushing
ΣForces = ΣF = ma
Since you are pushing an object uphill, P is positive (in order to go up) and friction is negative (it is preventing you from pushing the box easily)
Since a = 0 for (a), ma = 0
Therefore, P - f = ma
P - f = 0
P = f
For Part (b)
P - f = ma
P = ma + f
P = (20 kg)(0.75 m/s^2) + (whatever friction is on an inclined plane)
P = the sum of m*a and friction
The difference of your two answers is roughly 15 N, which is what (20 kg)(0.75 m/s^2) equals. However, your answers don't seem logical to me. When an surface becomes inclined, it becomes harder for friction to grasp the object (like in the movies when a ship sinks?), and therefore it should decrease.
But, whatever. Maybe I'm mistaken.
Σ=Sum of
f = friction
P=Force of pushing
ΣForces = ΣF = ma
Since you are pushing an object uphill, P is positive (in order to go up) and friction is negative (it is preventing you from pushing the box easily)
Since a = 0 for (a), ma = 0
Therefore, P - f = ma
P - f = 0
P = f
For Part (b)
P - f = ma
P = ma + f
P = (20 kg)(0.75 m/s^2) + (whatever friction is on an inclined plane)
P = the sum of m*a and friction
The difference of your two answers is roughly 15 N, which is what (20 kg)(0.75 m/s^2) equals. However, your answers don't seem logical to me. When an surface becomes inclined, it becomes harder for friction to grasp the object (like in the movies when a ship sinks?), and therefore it should decrease.
But, whatever. Maybe I'm mistaken.