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Question

wo forces are applied to a car in an effort to accelerate it, as shown below. The first force, F1 = 392 N, is applied at an angle α = 35° to the forward dashed line. The second force, F2 = 508 N, is applied at an angle β = 10° to the forward dashed line. 



(a) What is the resultant of these two forces? 
N 

(b) If the car has a mass of 2950 kg, what acceleration does it have? (Disregard friction.) 
m/s2

Answer 1

(a)

First Force F₁  = 392 N and at an angle  α = 35° .

So x - component of the force is F_1x = F₁ cos α .

F_1x = 392 cos 35 

F_1x = 392 (0.819)

F_1x = 321.10

y - component of the force is F_1y = F₁ sin α .

F_1y = 392 sin 35

F_1y =  392 (0.573)

F_1y = 224.84 

Second Force F₂  = 508 N N and at an angle  β = 10° .

So x - component of the force is F_2x =  F₂ cos  β .

F_2x  = 508 cos 10 

F_2x =  508 (0.9848)

F_2x  = 500.28

y - component of the force is F_2y =F₂  sin β .

F_2y = 508 sin 10

F_2y =  508 (0.1736)

F_2y = 88.213 

So the resultant force components are  is F_x =  F_1x + F_2x  and  F_y  =  F_1y + F_2y 

F_x = 321.10 + 500.28 

F_x = 821.38 

F_y  =  F_1y + F_2y

F_y  =  224.84  + 88.213 

F_y  = 313.053 

The resultant force is F  = √ [ (F_x)² + (F_y)² ] 

 F  = √ [ (88.213 )² + (313.053 )² ] 

F = 325.19 N .

So the resultant force is 325.19 N .

Answer 2

(b)

Mass of the car is 2950 kg .

Acceleration of the car be found using newton's law F = ma 

resultant force of the car is 325.19 N .                    [ From (a) ]

a = F/m 

a = 325.19 /2950 

a = 0.11 m/s²

So the acceleration of car is 0.11 m/s² .