Respuesta :
Answer:
F = 96.11 N
friction force =33.48 N
Explanation:
given,
mass of block M = 7.50 kg
acceleration = 3 m/s²
µ s = 0.443 and µ k = 0.312
angle with the horizontal = 25°
the component of force along the incline is F cos theta
normal reaction ,
[tex]N = mg cos \theta + F sin \theta[/tex]
[tex]N = 7.5\times 9.81\times cos25^0 + F sin 25^0[/tex]
N = 66.68+ .423 F
for the mass
[tex]F cos \theta - mg sin \theta - \mu_k\ N = ma[/tex]
[tex]F cos 25^0 - 7.5\times 9.81\times sin 25^0 - 0.312\times ( 66.68+ .423 F) = 7.5\times 3[/tex]
0.906 F - 0.132 F - 31.09 - 20.8 = 22.5
0.774F = 74.39
F = 96.11 N
normal force ,
N = 66.68 + .423×96.11
=107.33 N
friction force = .312 × N
= 0.312× 107.33
friction force =33.48 N
The friction force is the megnitude of force acting on a body in order to oppose the motion of the body. The megnitude of force acting on the body will be 96.11 N.
what is friction force?
The friction force is the megnitude of force acting on the body in order to oppose the motion of the body.
Friction is of two types one is static friction and another is kinetic friction
The static friction is act on the body when the body is just about to move.it is denoted by Fs. It is the product of static friction coefficient and normal reaction acting.
given,
coefficient of static friction = 0.443
coffiecient of kinetic friction = 0.312
[tex]\theta = 25^o[/tex]
F =?
N is the normal reaction act on the body
y component of the force component is given
[tex]\rm {N = mgcos\theta+fsin\theta}[/tex]
[tex]\rm{N = 7.5 \TIMES 9.81cos25^o+Fsin25^o}[/tex]----------1
Y component of the force component is given
[tex]Fcos\theta-mgsin\theta-u_kN = ma[/tex]
[tex]Fcos\25^0-7.5\times 9.81sin25^0-0.312(\rm{7.5 \TIMES 9.81cos25^o+Fsin25^o})= 7.5\times 3[/tex]
[tex]0.906 F-0.312 F -31.09 -20.8= 22.5[/tex]
[tex]0.774F = 74.39 \\\\\\\rm {F = 96.11 N}[/tex]
Hence the megnitude of force acting on the body will be 96.11 N.
To know more about friction force refer to the link;
https://brainly.com/question/1714663
