Question

Compute the horizontal force P required to prevent the block from sliding down the plane for the 100 lb block shown. Assume the coefficient of static friction to be 0.65.

Answer 1

canThe first First step we need to do is to make the decomposition of the vectors W and P.

Both will have a component perpendicular and parallel to the plane. The perpendicular will be used to calculate the maximum static friction force, and the horizontal will be used to find the P required to prevent the block from sliding.

From the sketch, we are able to define both, parallel and perpendicular components of P, as it follows:

`\begin{gathered} P_{\text{//}}=P\cos (30\degree)=\frac{P\sqrt[]{3}}{2} \\ P_(perp)=P\sin (30\degree)=(P)/(2) \end{gathered}`

Now, we can do the same for W.

And for W we also can provide the two components as follows:

`\begin{gathered} W_(//)=W\sin (30\degree)=(W)/(2) \\ W_{\text{perp}}=W\cos (30\degree)=\frac{W\sqrt[]{3}}{2} \end{gathered}`

Now, we can elaborate on both equations: one for the perpendicular direction and the other for parallel. In the perpendicular direction, we have a component of W, one component of P, and the normal force N. Because the block is going to move, or change its movement along this direction, the sum of the forces pointing upwards must be equal to the sum of the forces pointing downwards. From this, we can write the following:

`\begin{gathered} N=P_{\text{perp}}+W_{\text{perp}} \\ N=(P)/(2)+\frac{W\sqrt[]{3}}{2}=\frac{P+W\sqrt[]{3}}{2} \end{gathered}`

Now, for the horizontal, we have the P component to the right and the W component to the left. If we imagine the block is almost sliding. We can write the following equation, from the premise the forces will cancel each other just like the perpendicular case:

`\begin{gathered} P_(//)+F_(\mu)=W_(//) \\ \frac{P\sqrt[]{3}}{2}+N*\mu_(static)=(W)/(2) \end{gathered}`

Here it is used the fact that the friction force is equal to the multiplication of the coefficient of static friction by the normal force. Here we assumed also that the friction is maximum because the block is on the verge of motion downwards, and for this reason, the Friction is upwards, with the P component.

Now, substituting N and the coefficient, we find:

`\begin{gathered} \frac{P\sqrt[]{3}}{2}+\frac{P+100\sqrt[]{3}}{2}0.65=(100)/(2)=50 \\ \frac{P(\sqrt[]{3}+0.65)+65\sqrt[]{3}}{2}=50 \\ P(\sqrt[]{3}+0.65)+65\sqrt[]{3}=100 \\ P(\sqrt[]{3}+0.65)=100-65\sqrt[]{3} \\ P=\frac{100-65\sqrt[]{3}}{\sqrt[]{3}+0.65}\cong(100-65*1.732)/(1.732+0.65)=(100-112.58)/(2.382) \\ P=-(12.58)/(2.382)\cong-5.275\text{lbf} \end{gathered}`

From this, we can see that the force P made to the left with an intensity equal to -5.275 lbf will bring the block on the verge of motion downwards. If we consider that P is strong enough to make it almost move upwards, it is, the Normal Force will be downwards, we can remake the calculation as it follows:

`\begin{gathered} P_(//)=W_(//)+F_(\mu) \\ \frac{P\sqrt[]{3}}{2}=(W)/(2)+N*\mu_(static) \end{gathered}`

And substituting values, we have:

`\begin{gathered} \frac{P\sqrt[]{3}}{2}=50+\frac{P+100\sqrt[]{3}}{2}0.65 \\ \frac{P(\sqrt[]{3}-0.65)}{2}=50+50\sqrt[]{3}*0.65 \\ P=\frac{2}{\sqrt[]{3}-0.65}*50(1+\sqrt[]{3}*0.65)\cong196.46 \end{gathered}`

From this, we know that the max value for P, where the block will not slide is going to be 196.46 lbf to the right.