Question

Cherry and Cherra are pulling a box. Cherry pulls with 120 Newtons of force at 30° while Cheera pulls with 100 Newtons of force at 45° as shown. What is the combined force, and its direction?

a) What is Cherry’s Vector x and y?

b) What is Cherra’s Vector x and y?

c) Add the vectors of Cherry and Cherra.

d) What is the final direction?

Answer 1

Here the picture used for this task:

(a) We can find Cherry’s Vector x and y as follows:

$$F_{x, Cherry}=F_{Cherry}\cdot cos\theta=120\ N\cdot cos30^{\circ}=104\ N,$$ $$F_{y, Cherry}=F_{Cherry}\cdot sin\theta=120\ N\cdot sin30^{\circ}=60\ N.$$

(b) We can find Cherra’s Vector x and y as follows:

$$F_{x, Cherra}=F_{Cherra}\cdot cos\theta=100\ N\cdot cos(-45^{\circ})=70.71\ N,$$ $$F_{y, Cherra}=F_{Cherra}\cdot sin\theta=100\ N\cdot sin(-45^{\circ})=-70.71\ N.$$

(c) Let's first add the vectors of Cherry and Cherra:

$$F_{x,res}=F_{x, Cherry}+F_{x, Cherra}=104\ N+70.71\ N=174.71\ N,$$ $$F_{y,res}=F_{y, Cherry}+F_{y, Cherra}=60\ N+(-70.71\ N)=-10.71\ N.$$

Finally, we can find the magnitude of the combined force from the Pythagorean theorem:

$$F=\sqrt{(F_{x,res})^2+(F_{y,res})^2}=\sqrt{(174.71\ N)^2+(-10.71\ N)^2}=175\ N.$$

(d) We can find the direction of the combined force from the geometry:

$$\theta=tan^{-1}(\dfrac{F_{y,res}}{F_{x,res}}),$$ $$\theta=tan^{-1}(\dfrac{-10.71\ N}{174.71\ N})=-3.5^{\circ}.$$

The combined force directed at $3.5^{\circ}$ below the positive $x$-axis.