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?

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

asked Nov 6, 2014 in PHYSICS

(a)

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

So x - component of the force is F1x = F1 cos α .

F1x = 392 cos 35

F1x = 392 (0.819)

F1x = 321.10

y - component of the force is F1y = F1 sin α .

F1y = 392 sin 35

F1y =  392 (0.573)

F1y = 224.84

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

So x - component of the force is F2x =  F2 cos  β .

F2x  = 508 cos 10

F2x =  508 (0.9848)

F2x  = 500.28

y - component of the force is F2y =F2  sin β .

F2y = 508 sin 10

F2y =  508 (0.1736)

F2y = 88.213

So the resultant force components are  is F=  F1x F2x  and  F =  F1y F2y

F= 321.10 + 500.28

Fx = 821.38

F =  F1y F2y

F  224.84  + 88.213

F = 313.053

The resultant force is F  = √ [ (Fx)² + (Fy)² ]

F  = √ [ (88.213 )² + (313.053 )² ]

F = 325.19 N .

So the resultant force is 325.19 N .

(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 .