An inclined plane makes an angle of 30 degrees with the horizontal. Find the constant force, applied parallel to the plane, required to cause a 15 kg box to slide (a) up the plane with acceleration 1.2 m/s^2 and (b) down the incline with acceleration 1.2 m/s^2. Neglect friction forces.
The answer was (a) 91.5N and (b) 55.5N, but how do I get those answers?
The answer was (a) 91.5N and (b) 55.5N, but how do I get those answers?
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(a) By F(net) = F(applied) -F(gravity)
=>ma = F(a) - mgsinθ
=>F(a) = m(a+gsinθ)
=>F(a) = 15(1.2 + 9.8 x sin30*)
=>F(a) = 91.50 N
(b) By F(net) = F(applied) + F(gravity)
=>ma = F(a) + mgsinθ
=>F(a) = m(a-gsinθ)
=>F(a) = 15(1.2-9.8 x sin30*)
=>F(a) = -55.50 N [-ve just indicating the direction of the force is against the direction of motion]
=>ma = F(a) - mgsinθ
=>F(a) = m(a+gsinθ)
=>F(a) = 15(1.2 + 9.8 x sin30*)
=>F(a) = 91.50 N
(b) By F(net) = F(applied) + F(gravity)
=>ma = F(a) + mgsinθ
=>F(a) = m(a-gsinθ)
=>F(a) = 15(1.2-9.8 x sin30*)
=>F(a) = -55.50 N [-ve just indicating the direction of the force is against the direction of motion]