Answer to Question #200650 in Mechanics | Relativity for Master j

Question

A 50 kg wooden box is pulled up a plane inclined at 30^o to the horizontal with an acceleration of 5ms^-2. The pulling force, P, is applied parallel to the plane and the coefficient of kinetic friction between the box and the plane is 0.3.

i. Draw a diagram showing all the forces acting on the wooden box.

ii. Find the normal reaction, N_A

iii. Find the pulling force, P.

iv. If the wooden box was pulled with a constant velocity, determine the pulling force that would be required to pull the box up the incline

Answer 1

Explanations & Calculations

• Considering the equilibrium perpendicular to the plane,

$\qquad\qquad \begin{aligned} \small N_A-mg\cos 30&=\small 0\\ \small N_A&=\small 50kg\times9.8ms^{-2}\times\cos30\\ &=\small \bold{424.35\,N} \end{aligned}$

• Perpendicular to the plane, there is no acceleration hence no net force. All the forces are balanced out.

• By applying Newton's second law on the block for its motion upwards the incline, the pulling force could be calculated.

$\qquad\qquad \begin{aligned} \small P-f-mg\sin30&=\small ma\\ \small P&=\small f+mg\sin30+ma \end{aligned}$

• It is seen that it is needed to assess the frictional force that is applied on the block during the motion before we can calculate the pulling force.

$\qquad\qquad \begin{aligned} \small f&=\small \mu R=0.3\times424.35N\\ &=\small 127.31\,N \end{aligned}$

• Then,

$\qquad\qquad \begin{aligned} \small P&=\small 127.31+(50\times9.8\times\sin30)+(50kg\times5\,ms^{-2})\\ &=\small \bold{622.31\,N} \end{aligned}$

• At constant speed, there exists no acceleration hence no net force (reversed Newton's second)
• Then, forces in all directions are balanced.
• Therefore,

$\qquad\qquad \begin{aligned} \small P_1&=\small f+mg\sin30+0\\ &=\small \bold{372.31\,N} \end{aligned}$