A lorry of mass 1 tonne travels at a constant velocity up a slope inclined at 5degrees to the horizontal. The lorry experiences a resistive force of 1200N. Find the magnitude of the driving force of the lorry's engine.
I guess I have to use perpendicular/parallel forces but I'm confused as to whether I multiply 1 by 9.8? Would it be 1200+9.8sin5?
I guess I have to use perpendicular/parallel forces but I'm confused as to whether I multiply 1 by 9.8? Would it be 1200+9.8sin5?
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Constant velocity means the forces balance.
The weight is 1000 kg x 9.8 N/kg = 9.8 kN
the component of that pointing down the slope is
F = 9.8 kN sin 5 = 0.854 kN or 854 N
"The lorry experiences a resistive force of 1200N" ? this is open to interpretation.
One guess is that that is the friction and air resistance force it has on level ground. So in that case the engine has to overcome that PLUS the component of the weight, for a total of F = 1200 + 854 = 2054 N
The weight is 1000 kg x 9.8 N/kg = 9.8 kN
the component of that pointing down the slope is
F = 9.8 kN sin 5 = 0.854 kN or 854 N
"The lorry experiences a resistive force of 1200N" ? this is open to interpretation.
One guess is that that is the friction and air resistance force it has on level ground. So in that case the engine has to overcome that PLUS the component of the weight, for a total of F = 1200 + 854 = 2054 N