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Physics Question forces

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a tow of mass 1500kg is towing a car of mass 1000kg the horizontal force exerted on the car though the tow bar is 1000N this whole system as an acceleration of 0.5 m/s/s

Calculate
i)Friction on the towed car due to the road
ii)forward tractive force of the tow truck given that the friction opposing th tow truck iis 750 N


I want the working
the Answers given are
i)500N
ii)1500N
 
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AoA! :)
(i) Resultant force acting on the car, F = ma
=> F = (1000 x 0.5) N = 500 N.
Resultant force = applying force - opposing force(a.k.a. friction)
=> Friction = (1000 - 500) N = 500 N.
(ii) Resultant force acting on the tow truck = ma = (1500 x 0.5) = 750 N.
Apply the same formula: Resultant force = applying force - opposing force(a.k.a. friction)
So forward tractive force = (750 + 750) N = 1500 N.
Hope it helped. :)
 
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If it's the definition you require then here you go: The Principle of Moments states that for a body in equilibrium,the sum of clockwise moments about a pivot is equal to the sum of anticlockwise moments about the same pivot.
 
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abcde said:
AoA! :)
(i) Resultant force acting on the car, F = ma
=> F = (1000 x 0.5) N = 500 N.
Resultant force = applying force - opposing force(a.k.a. friction)
=> Friction = (1000 - 500) N = 500 N.
(ii) Resultant force acting on the tow truck = ma = (1500 x 0.5) = 750 N.
Apply the same formula: Resultant force = applying force - opposing force(a.k.a. friction)
So forward tractive force = (750 + 750) N = 1500 N.
Hope it helped. :)

in the 1st why didn't you applied f (friction) - F (force) = m x a ??
 
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Hey wouldnt we also count the 1000 N while calculating the forward tractive force????
 
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not really as it does not require u to answer that.

and @tanseer Force applied - friction = mass x acceleration is the formula given in the books so we have no choice. ;p
 
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