Statics 101 – Friction

A block is connected to a cable that runs over a frictionless pulley and connects to the right end of a beam that the block rests on, as shown below. If the weight of the beam is negligible compared to the weight of the block, find the minimum coefficient of friction between the block and the beam in terms of θ to prevent sliding.

Solution:

We begin by drawing a free body diagram of the beam with the block resting on it.

The tension in the cable is assumed to be uniform. Consider moment equilibrium about point a.

∑M z− po int a=TL+TL sinθ

2−WL

2=0→T= W

2+sin θ(101-1)

Now we draw a free body diagram of the block alone.

The friction force f is set equal to μN because we assume the block to be on the verge of slipping. Using the new free body diagram, we consider force equilibrium.

∑ F x=T cosθ−μN=0→ μ=T cosθ /N(101-2)

∑ F y=N−W+T sin θ=0→N=W−T sin θ(101-3)

Combining the three equations we can solve for μ.

μ=12

cosθ(101-4)

Author: Zaworski, J. R. Subject: StaticsTopic: FrictionProblem Length: 15 minutesGrading: Hand gradedPresentation Format: Text and figure with problem statementOther:

Statics 102 – Friction

A ladder rests against a wall. The coefficients of friction between the ladder and the wall and between the ladder and the floor are both equal to the same value μ=0.5. Find the angle θ at which the ladder will begin to slip.

Solution:

We start with a free body diagram of the ladder.

The friction forces are both set equal to μN because the ladder is on the verge of slipping. Consider moment equilibrium about point 1 where the ladder contacts the ground, taking the length of the ladder to be L.

∑M z−po int 1=WLcosθ

2−N 2 Lsin θ−0 .5N2 L cosθ=0

Next, let’s consider force equilibrium.

∑ F x=N2−0 .5N 1=0

∑ F y=N1+0. 5N2−W=0

Combining these three equations together gives

WLcosθ2

−0 . 4WL sin θ−0 .2WLcosθ=0

Next, we divide through by WL and replace sin by √1−cos2θ

to get

0 .5cosθ−0.4 √1−cos2θ−0 .2cos θ=0

This can be rearranged as

0 . 4√1−cos2θ=0 .3 cosθ

Squaring both sides gives

0 .16(1−cos2θ )=0 .09 cos2θ

Solving for gives

θ=36 .9°

Author: Kennedy, T. C.Subject: StaticsTopic: FrictionProblem Length: 20 minutesGrading: Hand gradedPresentation Format: Text and figure with problem statementOther:

Statics 103 – Friction

Three identical cylinders are stacked as shown. Find the minimum coefficient of friction μ1 between the cylinders and the minimum coefficient of friction μ2 between the cylinder and the floor to prevent collapse.

Solution:

We start with a free body diagram of the top cylinder taking into account symmetry in expressing the forces and assuming that the cylinders are on the verge of slipping. Note that if we draw imaginary lines connecting the centers of the circles, we get an equilateral triangle with 60° angles at the vertices.

Moment equilibrium and force equilibrium in the x-direction are already satisfied. Considering force equilibrium in the y-direction gives

∑ F y=2N1sin 60 °+2μ1N1 cos60°−W=0→N1=W

√3+μ1

Next, we draw a free body diagram of the left cylinder on the bottom.

Moment equilibrium about the center of the cylinder with radius r gives

∑M z−center=rμ2N2−rμ1N1=0→N2=μ1

μ2N 1=

μ1

μ2

W√3+μ1

Considering force equilibrium in the x-direction gives

∑ F x=μ2N 2−N1cos 60°+ μ1N 1 sin 60°= 0→μ1−12+μ1

√32

=0

Solving for μ1 gives

μ1=0 . 268

Considering force equilibrium in the y-direction gives

∑ F y=N2−N1sin 60 °−μ1N1 cos 60°−W=0→μ1

μ2N1−

√32N 1−

12μ1N1−W=0

Solving this for μ2 gives

μ2=μ1

√32 +

μ1

2 +WN1

Substituting the value for μ1 and the expression for N1 above, we get μ2 as

μ2=0 . 089

Author: Kennedy, T. C.Subject: StaticsTopic: FrictionProblem Length: 25 minutesGrading: Hand gradedPresentation Format: Text and figure with problem statementOther:

Statics 104 – Moment equilibrium and tipping

A mischievous student wants to tip over a sleeping 900lb cow. He can apply a horizontal force of 50lb as shown. If his friends are equally strong, how many of them will he need to get the job done.

Solution:

We begin by drawing a free body diagram of the cow.

Considering moment equilibrium about point b we get

∑M z−po int b=900 lb(1 ft )−F (4 .5 ft )−Ray(2 ft )=0

Solving this for Ray gives

Ray=450 lb−2. 25 F

We observe that Ray becomes negative when F becomes greater than 200lb. The vertical reaction force from the ground can never be negative. Therefore, F=200lb is the tipping point for the cow. Since one student can only apply 50lb, it will take four of them to get F up to 200lb. Therefore, the student will need to recruit three of his friends.

Author: Kennedy, T. C.Subject: Statics

Topic: Moment equilibrium and tipping Problem Length: 8 minutes Grading: Hand graded Presentation Format: Text and figure with problem statement Other:

Statics 105 – Friction

A yo-yo (with inner radius ri=0.5in and outer radius ro=1.5in) rests against a wall while its string is attached to the top of the wall as shown. Find the minimum coefficient of friction to prevent slip.

Solution:

We begin by drawing a free body diagram of the yo-yo.

Considering force equilibrium in the x-direction gives

∑ F x=N−T sin 30°=0→N=T /2

Considering moment equilibrium about the center gives

∑M z−center=Tri−μNr o=0→μ=TriNro

Substituting the values for ri and ro and N=T/2 gives μ as

μ=0 .667

Author: Kennedy, T. C.Subject: StaticsTopic: FrictionProblem Length: 8 minutesGrading: Hand gradedPresentation Format: Text and figure with problem statementOther:

Statics 106 – Friction

A block of weight W sits atop a cylinder with radius r. A string has one end attached to the block and the other end attached to a wall as shown below. The wheel is pinned at its center. What minimum moment M must be applied to the cylinder to cause it to rotate if the coefficient of friction between the block and the cylinder is μ?

Solution:

We assume that the block is on the verge of slipping so that the friction force between the block and cylinder is μ times the normal force. If we draw a free body diagram of the block and consider force equilibrium, we get

∑ F x=μN−T cosθ=0

∑ F y=N−W+T sin θ=0

Solving for T and N gives

N= W1+μ tan θ

Drawing a free body diagram of the cylinder and considering moment equilibrium about its center gives

∑M z−center=−M+μ Nr=0

Solving for M gives

M= μWr1+μ tan θ

Author: Kennedy, T. C.Subject: StaticsTopic: FrictionProblem Length: 8 minutesGrading: Hand gradedPresentation Format: Text and figure with problem statementOther:

Statics 107 – Force Equilibrium

A truck is stuck in mud. The passengers first attempt to dislodge it by pulling on a rope attached to the front bumper. After this fails, they try again with the end of the rope attached to a nearby tree before looping around the bumper. Ignoring friction between the rope and the bumper, compare the net pulling force on the truck between the first attempt and the second.

Solution:

We begin by drawing a free body diagram of the front bumper and rope for the first attempt.

Considering force equilibrium in the y-direction gives

∑ F y=T−R=0

Solving for the reaction force R gives the pulling force on the truck asR=T

Now we draw a free body diagram of the bumper and rope for the second attempt.

Considering force equilibrium in the x-direction and y-direction gives

∑ F x=Rx−T sin θ=0

∑ F y=T +T cosθ−R y=0

Solving for the components of the reaction force gives

Rx=T sin θand

R y=T (1+cosθ )

Calculating the resultant of the components gives the net pulling force as

R=√Rx2+R y2=√(T sin θ )2+[T (1+cosθ )]2=T √2(1+cosθ )

The pulling force in the second attempt ranges between 1.414 and 2 times the pulling force in the first attempt, depending on the value of θ.

Author: Kennedy, T. C.Subject: StaticsTopic: Force equilibriumProblem Length: 8 minutesGrading: Hand gradedPresentation Format: Text and figure with problem statementOther:

Statics 108 – Moment Equilibrium

A person tries to open a stuck bottle cap using a nutcracker as shown below. A normal force F1=10lb is applied to the lower arm, and a normal force F2=5lb is applied to the upper arm in an attempt to apply a counter-clockwise torque to the cap. The center of the cap is rigidly restrained. There is sufficient friction to prevent slip between the cap and the arms. Why will this not work for the position and forces shown?

Solution:

We begin by drawing free body diagrams of the two arms.

Considering moment equilibrium about points c and d gives

∑M z−po int c=10 lb (4 .67 in−1 .5 in )−Racos30 °(1.5 in )=0→Ra=24 .4 lb

∑M z− po int d=−5 lb(4 .67 in−1.5 in )+Rbcos 30° (1 .5 in )=0→Rb=12.2 lb

The forces Ra and Rb are the end forces in the cross member and should be equal. The fact that they are not tells us that the configuration in the figure is not a possible equilibrium state, and this technique will not work as shown.

Author: Kennedy, T. C.Subject: StaticsTopic: Force equilibriumProblem Length: 10 minutesGrading: Hand gradedPresentation Format: Text and figure with problem statementOther:

Statics 109 – Force and Moment Equilibrium

A force P is applied to the mechanism shown below. The thickness of each bar is 2in. If the maximum allowable force in the pin at b is 100lb, find the maximum allowable value of P, assuming that the weights of the components are negligible compared to P..

Solution:

We begin by drawing a free body diagram of the entire assembly.

Considering moment equilibrium about point d gives

∑M z−po int d=P(2 in )−R1 y(32 in )=0→R1 y=P/16

Considering force equilibrium gives

∑ F x=Rdx−P=0→Rdx=P

∑ F y=R1 y−Rdy=0→Rdy=P/16

Now we draw a free body diagram of the bottom arm.

Considering moment equilibrium about point b gives

∑M z−po int b=−R2 y(12 in )− P16

(22 in)+P(8 in)=0→R2 y=0 .552P

Considering force equilibrium gives

∑ F x=P−Rbx=0→Rbx=P

∑ F y=Rby−R2 y−P/16=0→Rby=0 .615 P

The resultant force on the pin is

Rbnet=√Rbx2 +Rby2 =√P2+(0 . 615P )2=1.174 P

Setting this equal to the allowable value of 100lbs gives

P=85. 2 lb

Author: Kennedy, T. C.Subject: StaticsTopic: Force and Moment equilibriumProblem Length: 18 minutesGrading: Hand gradedPresentation Format: Text and figure with problem statementOther:

Statics 110 – Force and Moment Equilibrium in Three Dimensions

A 20” by 30” lid weighing 10lbs is held at an angle θ=30° by a force P normal to the lid as shown below. Determine the magnitude of P and the reaction forces in the hinges at the bottom corners.

Solution:

We begin by drawing a free body diagram of the lid assuming that there are no z-components of force in the hinges (What would happen if these were included?).

Considering moment equilibrium about the z-axis through point a gives

∑M z−po int a=P(20in )−10 lb(20 in /2)cos30 °=0→P=4 . 33lb

Considering moment equilibrium about the x-axis through point a gives

∑M x− po int a=Rby(30 in)+P cos30 ° (20in)−10lb (15 in )=0→Rby=2.5 lb

Considering moment equilibrium about the y-axis through point a gives

∑M y−po int a=Rbx (30 in)−P sin 30°(20 in)=0→Rbx=1. 44 lb

Considering force equilibrium in the x-direction gives

∑ F x=Rax+Rbx−P sin 30 °=0→Rax=0 .725 lb

Considering force equilibrium in the y-direction gives

∑ F y=Ray+Rby−10 lb+P cos30 °=0→Ray=3 .75lb

Author: Kennedy, T. C.Subject: StaticsTopic: Force and Moment equilibrium in Three DimensionsProblem Length: 15 minutesGrading: Hand gradedPresentation Format: Text and figure with problem statementOther: