Question

Daisy walks across a force platform, and forces exerted by her foot during a step are recorded. The peak velocity reaction force is 1200 (this force acts upward on Daisy). At the same instant, the frictional force is 200N(this force acts forward on Daisy).(a) how large is the resultant of these forces(b) What is the direction of the resultant forces?

Answer 1

Answer:

a) $F_r=1216.55\ N$

b) $\theta=80.54^{\circ}$

Explanation:

Given:

• force acting upward on the, $F_y=1200\ N$

• force acting forward on daisy, $F_x=200\ N$

a)

Now the resultant of these forces:

Since the forces are mutually perpendicular,

$F_r=\sqrt{F_x^2+F_y^2}$

$F_r=\sqrt{200^2+1200^2}$

$F_r=1216.55\ N$

b)

The direction of this force from the positive x-direction:

$\tan\theta=\frac{F_y}{F_x}$

$\tan\theta=\frac{1200}{200}$

$\theta=80.54^{\circ}$

Answer 2

Answer:

(a). The resultant of these forces is 1216.55 N.

(b).  The direction of the resultant forces is 80.53°.

Explanation:

Given that,

First force = 1200 N

Second force = 200 N

(a). We need to calculate the resultant of these forces

Using cosine law

$F=\sqrt{F_{1}^2+F_{2}^2+2F_{1}F_{2}\cos\theta}$

Put the value into the formula

$F=\sqrt{1200^2+200^2+2\times1200\times200\cos90}$

$F=\sqrt{1200^2+200^2}$

$F= 1216.55\ N$

The resultant of these forces is 1216.55 N.

(b). We need to calculate the direction of the resultant forces

Using formula of direction

$\tan\alpha=\dfrac{F_{1}}{F_{2}}$

Put the value into the formula

$\alpha=\tan^{-1}(\dfrac{1200}{200})$

$\alpha=80.53^{\circ}$

Hence, (a). The resultant of these forces is 1216.55 N.

(b).  The direction of the resultant forces is 80.53°.