MHB Forces of 8N: Find Angle & Mass of Particle

[Shah 72]

Two forces each of size 8N, have a resultant of 13N.

a) Find the angle between the forces

b) The two given forces of magnitude 8N act on a particle of mass m kg, which remains at rest on a horizontal surface with no friction. The normal contact force between the surface and the particle has magnitude 7N. Find m and the acute angle that one of the 8N forces makes with the surface.

 

[HOI]

What have you done on this and where do you have a problem? Have you drawn the "force parallelogram" showing the addition of the forces?

 

[Shah 72]

What have you done on this and where do you have a problem? Have you drawn the "force parallelogram" showing the addition of the forces?

Hi , thanks I got the answer.

 

[skeeter]

(a) angle between the two 8N forces is

$\theta = \pi - \arccos\left(\dfrac{2 \cdot 8^2 - 13^2}{2\cdot 8^2}\right)$, or $\theta = 2\arccos\left(\dfrac{6.5}{8}\right)$

(b) $7\,N = mg - 13\,N \implies mg = 20\,N \implies m = 2kg$ (using $g = 10 \, m/s^2$)

$\phi = \dfrac{\pi - \theta}{2}$

 

[Shah 72]

(a) angle between the two 8N forces is

$\theta = \pi - \arccos\left(\dfrac{2 \cdot 8^2 - 13^2}{2\cdot 8^2}\right)$, or $\theta = 2\arccos\left(\dfrac{6.5}{8}\right)$

(b) $7\,N = mg - 13\,N \implies mg = 20\,N \implies m = 2kg$ (using $g = 10 \, m/s^2$)

$\phi = \dfrac{\pi - \theta}{2}$

Thank you so much!