Question

A 10kg box is pulled along a rough surface with force 30N at a 30-degree angle.

(a) What are the horizontal and vertical components of this applied force?

(b) What is the force of gravity on the box?

(c) Normal Force?

(d) If the coefficient of the kinetic friction is .25, what is the net force and acceleration?

Answer 1

Part a Let us first draw a figure and analyse the direction of force with respect to horizontal

​​​​​Since , horizontal component of force F (H ) is given by

F(H) = F Cos(theta)

Where theta is the angle made by the force with horizontal direction i.e theta = 30°

And F = 30 N

Hence, F (H) = 30 Cos(30°)

= 30*0.866

= 25.98 N

Similarly, vertical component of the force F(V)is given by

F(V) = F sin(theta)

= 30*sin30°

= 30*0.5

= 15 N

Part b. The force of gravity F(G)on box will be vertically downwards equal to its weight given by

F (G) = mg

Since m = 10 kg and g = 9.8 m/s^{​​​​​​2}

Hence, F(G) = 10*9.8

= 98 N

Part c. The normal force is perpendicular to the two surfaces in contact, it is also called normal reaction and balances the weight of the body acting in downward direction, i.e it acts in vertically upward direction.

However the magnitude would be given by:

Normal force in such a case is given by

F(N) = F(G) - F(V)

= 98- 15

F(N) = 83 N

Further, this normal force would act in vertically upward direction.

Part d. Now, the frictional force F(F) and normal force F(N) are related as

F(F) = uF(N)

where u is the coefficient of kinetic friction.

u = 0.25

Hence F(F) = 0.25*83

= 20.25 N

This frictional force will act in direction opposite to horizontal component of the external force applied since it would oppose the motion of the crate. Hence, net force would be given by

F(net) = F(H) - F(F)

= 25.98- 20.25 N

= 5.23 N

And accordingly acceleration a produced in the body in the direction of net force is given by

a = F(net)/M

= 5.23/10

= 0.523 m/s^{​​​​​​2}