Answer:
a) Normal = 50.3 N
b) friction = 26.16 N
c) acceleration = 2.14 [tex]\frac{m}{s^2}[/tex]
Explanation:
Let's start by drawing the free-body diagram of all forces acting on the box (see attached image). Apart from the force F acting at 24 degrees from the horizontal, we have the weight (w) of the box (force exerted by gravitational pulling of the Earth), the normal (n) exerted by the floor on the box, and the force of friction (f ) with the floor, acting opposite to the box's displacement.
Notice that the applied force F acts at an angle, therefore, there is a vertical component of it (pointing up) which is going to act in the opposite direction to the weight, thus making the box somehow "lighter", so the normal exerted by the floor on the box will be in magnitude less than the weight of the box. The actual vertical component of the force F will be given by the product of the magnitude of F times the sine of the angle. This quantity will then subtract from the weight to give us the magnitude of the normal force that the floor will be applying to compensate and prevent the box from moving vertically:
[tex]n=w-F\,sin(24^o)\\n=7\,kg\,*g-45\,N\,sin(24^o)\\n=50.2968\,N[/tex]
which we can round to 50.3 N
Now we need to find the force of friction (f ), which is defined as the product of the coefficient of friction times the normal force (found above):
[tex]f=\mu\,n\\f=0.52\,*\,50.3 \,N\\f=26.156 \,N[/tex] which we can round to 26.16 N
Notice now that the box is not moving vertically (therefore there is no vertical net force, but it is moving horizontally. The horizontal movement is due to the horizontal component (F times the cosine of the angle) of the applied force F, minus the friction force (f ), and this difference will give us the actual net force acting on the box:
[tex]F_{net}=F\,cos(24^o)-f\\F_{net}=45\,cos(24^o)-26.16\,N\\F_{net}=14.9495\,N[/tex]
which we can round to 14.95 N
Now we can apply Newton's second law to this in order to find the acceleration of the box:
[tex]F_{net}=m\,a\\14.95\,N=7\,kg\,a\\a=\frac{14.95}{7} \frac{m}{s^2} \\a=2.1357\,\frac{m}{s^2}[/tex]
which we can round to 2.14 [tex]\frac{m}{s^2}[/tex].
Notice that we know the answer comes in [tex]\frac{m}{s^2}[/tex], since we included all SI units in our equation.
