What is the magnitude and direction of the resultant force?

Question

1. A 110-N force and a 55-N force both act on an object point P. The 110-N force acts at 90 degree. the 55-N force acts at 0 degree. What is the magnitude and direction of the resultant force?

2. A motorboat travels at 8.5 m/s. It heads straight across a river 110 m wide.

a. If the water flows downstream at a rate 3.8 m/s, what is the boat's resultant velocity?

b. How long does it take the boat to reach the opposite shore?

I really need it now. For the reason that our finals is tomorrow morning.

I need to know the answer so that it can help me understand it.

any response will be greatly appreciated.

Answer 1

1. Let us assume that the forces are in XY plane and the angles given are from x axis.

110N force acts at 90 deg. This means it is along Y axis.

55N force acts at 0 deg. This means it is along X axis.

X component of resultant force = Fx = 55N

Y component of resultant force = Fy = 110N

Magnitude of the resultant force = √(Fx^2 + Fy^2) = √(55^2 + 110^2) N

= √(3025 + 12100) N

= √(15125) N

= 123N

Let the resultant force makes angle θ with X axis.

Then tanθ= Fy/Fx

Or tanθ = 110/55 = 2

θ = atan(2) = 63.4 deg

Ans: Magnitude of resultant force = 123 N, direction = at angle 63.4 deg



2a)Let v = motorboat's velocity

u = water's velocity

v = 8.5 m/s

u = 3.8 m/s

u and v are at 90 deg. Therefore, the boat's resultant velocity = √(v^2 + u^2) = √(8.5^2 + 3.8^2) m/s

= √(72.25 + 14.44) m/s

= √86.69 m/s

= 9.3 m/s

If the resultant velocity makes angle θ with the direction in which water flows, then

tanθ = v/u

Or tanθ = 8.5/3.8 = 2.24

Therefore, θ = atan(2.24) = 66 deg

Ans: 9.3 m/s at 66 deg from the direction of water flow.

b)Displacement along the width of the river = 110 m

v is along the width of the river and u is perpendicular to the width.

Therefore, time = 110m/v

= 110/8.5 sec

= 12.94 sec

Ans: 12.94 sec

Note: It will be good if you show some simple diagram depicting the problem.

Answer 2

♠ thus vector F1=(cos(90), sin(90))*110N, vector F2=(cos(0), sin(0))*55N;

therefore F=F1+F2 =(55, 110); |F|=√(55^2 +110^2) =55*√5 N;  

♣boat’s resultant velocity is a vector w=(3.8, 8.5)m/s;

time to reach the shore is t=110/8.5 =12.94s;