**Part (a) – Force P to just start the block to move up the incline**
The force P is pushing the block up the incline. The push is hard enough to overcome the maximum allowable friction causing an impending upward motion.

$\Sigma F_y = 0$

$N = 2225 \cos 45^\circ + P \sin 45^\circ$

$N = 1573.31 + 0.7071P$

$f = \mu N = 0.25(1573.31 + 0.7071P)$

$f = 393.33 + 0.1768P$

$\Sigma F_x = 0$

$P \cos 45^\circ = f + 2225 \sin 45^\circ$

$P \cos 45^\circ = (393.33 + 0.1768P) + 2225 \sin 45^\circ$

$0.5303P = 1966.64$

$P = 3708.55 \, \text{ N}$ *answer*

**Part (b) – Force P to just prevent the block to slide down the incline**

In this case, the force P is not pushing the block upward, it simply supports the block not to slide downward. Therefore, the total force that prevents the block from sliding down the plane is the sum of the component of P parallel to the incline and the upward friction force.

$\Sigma F_y = 0$

$N = 2225 \cos 45^\circ + P \sin 45^\circ$

$N = 1573.31 + 0.7071P$

$f = \mu N = 0.25(1573.31 + 0.7071P)$

$f = 393.33 + 0.1768P$

$\Sigma F_x = 0$

$P \cos 45^\circ + f = 2225 \sin 45^\circ$

$P \cos 45^\circ + (393.33 + 0.1768P) = 2225 \sin 45^\circ$

$0.8839P = 1179.98$

$P = 1335 \, \text{ N}$ *answer*

**Part (c) – Force P = 1780 N**

If P_{x} = W_{x}, there will be no friction under the block. If P_{x} > W_{x}, friction is going downward to help W_{x} balance the P_{x}. If P_{x} < W_{x}, friction is going upward to help P_{x} balance the W_{x}. In this problem, the maximum available friction is not utilized by the system.

$W_x = 2225 \sin 45^\circ = 1573.31 ~ \text{N}$

$P_x = 1780 \cos 45^\circ = 1258.65 ~ \text{N}$

W_{x} > P_{x}, thus, f is upward.

$\Sigma F_x = 0$

$f + P_x = W_x$

$f + 1258.65 = 1573.31$

$f = 314.66 ~ \text{N upward}$ *answer*

d po kaya sa part c. mga boss goes down...kasi ang minimum force req'd from goin down is na-compute na which is 1335N...eh ung px ay 1258N lang....

yung 1335N boss ehh yung mismong force P hindi yan minimum dahil hindi naman dinerivative. Kaya po upward yung f para po i balance nya sa Wx. Kasi sa problem na yan Wx>Px. Sa part C kasi nag bigay sya ng ibang value ng P which is yung 1780 N. Take note boss yung 1258 ay Px which is yung x-component ng force P. Kaya magkaibang magkaiba talaga yan sila. So kapag mataas ang Wx ang f ay upward para nga ma i balance nya yung object. Kapag Px naman ang f ay downward. Magkaiba po ng scenario ang part B at part C.