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Questions: 1 2 3 4 5 6 7 8

Physics 1101 Static Equilibrium
  1. In the diagram below, three forces are applied to a 3-4-5 triangle. The forces are F1 = 91.7 N, F2 = 150 N, and F3 = 67.7 N. F3 is applied at the middle of side AB. (a) Find the net torque about point A. (b) Find the net torque about point B. (c) Find the net torque about point C.

    There are two methods for determining torque. Method A is to use τz = rFsinθ, where r is the distance from the pivot point to the point where the force F acts and θ is the angle between r and F. The sign of τz is found using the right-hand rule. Method B is to use τz = xFy - yFx, where (x, y) is the location of where the force is acting taken relative to the pivot point which is taken to be the origin (0, 0). Fx and Fy are the components of the force - careful attention must be paid to signs.

    Method A.

    First note that the interior angles of the triangle are α = tan-1(4/5) = 53.130°, and γ = tan-1 (3/4) = 36.870°. F2 makes an angle φ = 180° - 110° - γ = 16.870° with the vertical.

    (a)

    r (m) F (N) θ direction τz = rFsinθ (N-m)
    091.7- - 0
    5 150 110° CCW 704.769
    2 67.7 130° CW -103.722
    Total: 601.0

    (b)

    r (m) F (N) θ direction τz = rFsinθ (N-m)
    4 91.7 90° CCW 366.800
    3 150 110° + γ CCW 130.591
    2 67.7 50° CCW 103.722
    Total: 601.1

    (c)

    r (m) F (N) direction τz = rFsinθ (N-m)
    5 91.7 90° + α CCW 366.800
    0 150 - - 0
    3.60555 67.7 50° + 56.130° CCW 234.272
    Total: 601.1

    Method B:

    (a)

    x (m) y (m) Fx (N) Fy (N) τz = xFy - yFx (N-m)
    0 0 0 -91.7 0
    4 3 -150sinφ 150cosφ 704.769
    2 0 67.7cos50° -67.7sin50° -103.722
    total: 601.0

    (b)

    x (m) y (m) Fx (N) Fy (N) τz = xFy - yFx (N-m)
    -4 0 0 -91.7 103.722
    0 3 -150sinφ 150cosφ 130.591
    -2 0 67.7cos50° -67.7sin50° 366.80
    total: 601.1

    (c)

    x (m) y (m) Fx (N) Fy (N) τz = xFy - yFx (N-m)
    -4 -3 0 -91.7 366.800
    0 0 -150sinφ 150cosφ 0
    -2 -3 67.7cos50° -67.7sin50° 234.272
    total: 601.1

    Please note that the only reason the total torque at point A, B, and C are the same is because F1 + F2 + F3 = 0.

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  2. An L-shaped object of uniform density is hung over a nail so that it is free to pivot. What angle, θ, does the long side make with the vertical? The long side of the L-shaped object is twice as long as the short side?

    The problem mentioned that the object is free to pivot, to rotate. This indicates that we are dealing with a Static Equilibrium problem. We solve Static Equilibrium problems by sketching the extended free-body diagram, an FBD where the location of the all forces are indicated so that torques can be calculated. Then we determine the three equations necessary for static equilibrium, ΣFx = 0, ΣFy = 0, and Στz = 0.

    The forces that we know are working on the L-shaped object are a normal from the nail and the weight which acts from the centre of mass. Ordinarily, for complex shapes, we first determine the CM. However, in this case, it is easier to consider the two arms of the objects as being separate objects. The long arm will have a mass (2/3)mg and the short arm will be (1/3)mg. We do not have a simple method of figuring out which way the normal points. As with all pins, we consider it as two forces one vertical and one horizontal.

    ΣFx = 0 ΣFy = 0
    Nx = 0 Ny - (1/3)mg - (2/3)mg = 0

    These tell us the obvious, the normal has no horizontal component and that it supports the weight of the object.

    We will use Method A for the torques since that method is easiest to apply here. We will take the nail as the pivot point since this eliminates the torques from the nail.

    r F direction τz = rFsinθ
    0 Nx - - 0
    0 Ny - - 0
    L/2 (1/3)mg π+θ CW -mgLsin(π+θ)/6
    L (2/3)mg θ CCW 2mgLsinθ/3

    Since Στz = 0, the equation we get is

    -mgLsin(π+θ)/6 + 2mgLsinθ/3 = 0.

    Eliminating common terms and noting sin(π+θ) = cosθ, this becomes

    -cosθ/2 + 2sinθ = 0,

    or

    sinθ/cosθ = 1/4.

    Using the identity, tanθ = sinθ/cosθ, we thus have θ = tan-1(1/4) = 14.0°. The long side makes a 14.0° angle with the vertical.

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  3. A uniform 400 N boom is supported as shown in the figure below. Find the tension in the tie rope and the force exerted on the boon by the pin at P.

    The problem mentions forces and looking at the diagram shows that the object would rotate in the absence of any one of these forces. This indicates that we are dealing with a Static Equilibrium problem. We solve Static Equilibrium problems by sketching the extended free-body diagram, an FBD where the location of the all forces are indicated so that torques can be calculated. Then we determine the three equations necessary for static equilibrium, ΣFx = 0, ΣFy = 0, and Στz = 0.

    The forces that we know are working on the boom are a normal from the pin, the weight which acts from the centre of mass, and the two tensions. We are given T2. The CM is obviously at the centre of the boom. We do not have a simple method of figuring out which way the normal points, instead we consider it as two forces one vertical and one horizontal.

    ΣFx = 0 ΣFy = 0
    Px - T1 = 0 Py - mg - T2 = 0

    These tell us the obvious, that Px = T1 and Py = mg + T2 = 2400 N.

    We will use Method A for the torques since that method is easiest to apply here since the distances and angles are relatively easy to find. We will take the pin as the pivot point since this eliminates the torques from the pin.

    r F θ direction τz = rFsinθ
    0 Px - - 0
    0 Py - - 0
    L/2 W 40° CW -LWsin(40°)/2
    (3/4)L T1 50° CCW 3LT1sin(50°)/4
    L T2 40° CW -LT2sin(40°)

    Since Στz = 0, the equation we get is

    -WLsin(40°)/2 + 3LT1sin(50°)/4 - LT2sin(40°) = 0 .

    Eliminating L from the above and rearranging to get T1 by itself yields,

    T1 = {2[W + 2T2]sin(40°)} / 3sin(50°) .

    Using the values we are given, we find T1 = 2461 N.

    Since we know Px = T1, we also know the pin force is

    P = i2461 N + j2400 N.

    The magnitude of this force is P = [(Px)2 + (Py)2 ]½ = 3438 N. The force is directed at an angle θ = tan-1(Py/Px) = 44.3° to the horizontal. Note that the pin force is not pointed solely along the length of the boom as one might expect.

    Big Point To Remember: Pin forces are not always directed in the obvious direction.

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  4. In the figure below, a mass of 500 kg is held motionless in the air by a 120-kg boom and a rope. Find the tension in the rope. Find the force exerted on the boom by the pin at P. The angles are q = 30.0° and φ = 45.0°.

    The problem mentions forces and looking at the diagram shows that the object would rotate in the absence of any one of these forces. This indicates that we are dealing with a Static Equilibrium problem. We solve Static Equilibrium problems by sketching the extended free-body diagram, an FBD where the location of the all forces are indicated so that torques can be calculated. Then we determine the three equations necessary for static equilibrium, ΣFx = 0, ΣFy = 0, and Στz = 0.

    The forces that we know are working on the boom are a normal from the pin, the weight which acts from the centre of mass, and the two tensions. The CM is obviously at the centre of the boom. We do not have a simple method of figuring out which way the normal points, instead we consider it as two forces one vertical and one horizontal.

    Fx = 0 ΣFy = 0
    Px - T1cos(π/2-θ) = 0 Py - mg - T2 - T1sinθ = 0

    These tell us that Px = T1cosθ and Py = mg + T2 + T1sinθ. We are given the mass of the load so we know T2 = mloadg = 4905 N.

    We will use Method A for the torques since that method is easiest to apply here since the distances and angles are relatively easy to find. We will take the pin as the pivot point since this eliminates the torques from the pin.

    r F θ direction τz = rFsinθ
    0Px - - 0
    0 Py - - 0
    L/2 mg π/2 - φ CW -Lmgsin(π/2-φ)/2
    L T1 θ - φ CCW LT1sin(θ-φ)
    L T2 π/2 - φ CW -LT2sin(π/2-φ)

    Since Στz = 0, the equation we get is

    -Lmg[sin(π/2-φ)]/2 + LT1sin(θ-φ) - LT2sin(π/2-φ) = 0 .

    Eliminating L from the above, multiplying through by 2, and using the identity that sin(π/2-φ) = cosφ yields,

    -mgcosφ + 2T1sin(θ-φ) - 2T2cosφ = 0.

    We rearrange to get T1 by itself,

    T1 = (mg + 2T2)cosθ / 2sin(θ-φ) .

    Using the values we are given, and the value for T2, we find T1 = 15008 N.

    We have Px = T1cosθ = 12998 N. As well, Py = mg + T2 + T1sinθ = 13587 N. Thus we also know that the pin force is

    P = i12998 N + j13587 N.

    The magnitude of this force is P = [(Px)2 + (Py)2 ]½ = 1.88 × 104 N. The force is directed at an angle θ = tan-1(Py/Px) = 46.3° to the horizontal. Note that the pin force is not pointed solely along the length of the boom as one might expect.

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  5. A rectangular sign of mass 50.0 kg and width w = 5.00 m and l = length 4.00 m is hanging from a hinge and a rope as shown in the figure below. The rope makes and angle θ = 65.0° with the right wall.
    (a) Find the tension in the rope.
    (b) Find the horizontal and vertical components of the hinge force.

    The problem mentions forces and looking at the diagram shows that the object would rotate in the absence of any one of these forces. This indicates that we are dealing with a Static Equilibrium problem. We solve Static Equilibrium problems by sketching the extended free-body diagram, an FBD where the location of the all forces are indicated so that torques can be calculated. Then we determine the three equations necessary for static equilibrium, ΣFx = 0, ΣFy = 0, and Στz = 0.

    The forces that we know are working on the sign are a normal from the pin, the weight which acts from the centre of mass, and the tension. The CM is obviously at the centre of the sign. We do not have a simple method of figuring out which way the normal points, instead we consider it as two forces one vertical and one horizontal.

    ΣFx = 0 ΣFy = 0
    Px - Tsinθ = 0 Py - mg + Tcosθ = 0

    These tell us that Px = Tsinθ and Py = mg - Tcosθ.

    We will use Method B for the torques since that method is easiest to apply here since the location of the forces easy to find. We will take the pin as the pivot point since this eliminates the torques from the pin.

    x y Fx Fy τz = xFy - yFx
    0 0 Px Py 0
    w 0 Tsinθ Tcosθ wTcosθ
    w/2 -l/2 0 -mg -wmg/2

    Since Στz = 0, the equation we get is

    wTcosθ - wmg/2 = 0 .

    Eliminating w from the above and rearranging yields,

    T = mg / 2cosθ = 580.3 N .

    The force equations give Px = Tsinθ = 526 N and Py = mg - Tcosτq = 245.0 N.

    Thus the tension in the rope is 580 N and the horizontal and vertical components of the pin force are 526 N and 245 N respectively.

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  6. Determine the tension in the strings and the unknown angle θ. Each square has a side of length 32.0 cm. The object has a mass of 125 g.

    The problem mentions forces and looking at the diagram shows that the object would rotate in the absence of any one of these forces. This indicates that we are dealing with a Static Equilibrium problem. We solve Static Equilibrium problems by sketching the extended free-body diagram, an FBD where the location of the all forces are indicated so that torques can be calculated. Then we determine the three equations necessary for static equilibrium, ΣFx = 0, ΣFy = 0, and Στz = 0.

    The forces that we know are working on the sign are the two tensions and the weight which acts from the centre of mass of each piece. Each piece has one-third of the total mass and weight.

    ΣFx = 0 ΣFy = 0
    T1sinθ - T2sin(65°) = 0 T1cosτq + T2cos(65°) - Mg = 0

    These tell us that T1sinθ = T2sin(65) and T1cosθ + T2cos(65°) = Mg.

    We will use Method B for the torques since that method is easiest to apply here since the location of the forces easy to find. We will locate the pivot at the upper right corner because we have two unknowns there.

    x(m) y( m) Fx Fy τz = xFy - yFx
    0 0 T1sinθ T1cosθ 0
    -0.64 -0.32 -T2sin(65°) T2cos(65°) 0.64T2(65°) - 0.32T2sin(65°)
    +0.16 -0.48 0 -(1/3)Mg -(0.16/3)Mg
    -0.16 -0.16 0 -(1/3)Mg +(0.16/3)Mg
    -0.48 -0.48 0 -(1/3)Mg 0.16Mg

    Since Στz = 0, the equation we get is

    -0.64T2cos(65°)- 0.32T2sin(65°) + 0.16Mg = 0 .

    Rearranging the above yields the tension in the left string,

    T2 = 0.16Mg / [0.64cos(65°) + 0.32sin(65°)] = 0.3500 N .

    The force equations give

    T1sinθ = T2sin(65°) = 0.31725 N, and

    T1cosτ = Mg - T2cos(65°) = 1.07831 N.

    Taking the ratio of these two results we have sinθ/cosθ = 0.31725/1.07831 or tanθ = 0.2942. So the unknown angle is θ = 16.4°. Substituting the angle back into either of the two equations yields the tension in the right string T1 = 1.124 N.

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  7. The sign has a mass of 20.0 kg. The hinge is located at the bottom of the left side. Find the centre of mass. Determine the tension in the rope and the horizontal and vertical components of the hinge force. The length, l, is 12 cm.

    The problem mentions forces and looking at the diagram shows that the object would rotate in the absence of any one of these forces. This indicates that we are dealing with a Static Equilibrium problem. We solve Static Equilibrium problems by sketching the extended free-body diagram, an FBD where the location of the all forces are indicated so that torques can be calculated. Then we determine the three equations necessary for static equilibrium, ΣFx = 0, ΣFy = 0, and Στz = 0.

    The sign is complex so we break it up into smaller regularly-shaped pieces. The sign has a total area of 9l2. The top piece has an area of 5l2, therefore its mass is (5/9)M. The vertical piece has an area 4l2, therefore its mass is (4/9)M.

    The forces that we know are working on the sign are the tension, and the weight of each piece which acts from the centre of mass (geonmetric centre) of each piece, and the normal force from the hinge. Since we do not know the direction of the normal force, we show components.

    ΣFx = 0 ΣFy = 0
    -Hx + Tsin(40°) = 0 Hy + Tcos(40°) - Mg = 0

    These tell us that Hx = Tsin(40°) and Hy + Tcos(40°) = Mg.

    We will use Method B for the torques since that method is easiest to apply here since the location of the forces easy to find. We will locate the pivot at the hinge because we have two unknowns there.

    x y Fx Fy τz = xFy - yFx
    0 0 Hx Hy 0
    5l l Tsin(40°) Tcos(40°) 5lTcos(40°) - lTsin(40°)
    (5/2)l ½l 0 -(5/9)Mg -(25/18)lMg
    (5/2)l -2l 0 -(4/9)Mg -(20/18)lMg

    Since Στz = 0, the equation we get is

    5lTcos(40°) - lTsin(40°) - (45/18)lMg = 0 .

    Eliminating l and rearranging the above yields the tension in the rope,

    T = (5/2)Mg / [5cos(40°) - sin(40°)] = 153.9 N .

    The force equations give

    Hx = Tsin(40°) = 98.9 N , and

    Hy = Mg - Tcos(40°) = 78.3 N.

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  8. A ladder is propped against a wall making an angle with the floor. The wall is frictionless but the coefficients of friction for the floor are μs and μk respectively. Obtain an expression for the smallest that can be if the ladder is not to slip. Recall that tanθ = sinθ/cosθ.

    The problem mentions forces and looking at the diagram shows that the object would rotate in the absence of any one of these forces. This indicates that we are dealing with a Static Equilibrium problem. We solve Static Equilibrium problems by sketching the extended free-body diagram, an FBD where the location of the all forces are indicated so that torques can be calculated. Then we determine the three equations necessary for static equilibrium, ΣFx = 0, ΣFy = 0, and Στz = 0.

    The forces that we know are working on the ladder are the weight which acts from the centre of mass, the normal forces from the wall and floor, and friction. Since the ladder is not moving, we are dealing with static friction. Since we want the smallest angle, we are dealing with fs MAX. Since the ladder has a tendency to move to the left, friction points to the left.

    ΣFx = 0 ΣFy = 0
    fs MAX - Nw = 0 Nf - mg = 0

    These tell us that fs MAX = μNw and Nf = mg. We also know that fs MAX = μsNf where Nf is the normal force between the ladder and floor. As a result, we have fs MAX = μsmg. Hence Nw = μsmg as well.

    We will use Method A for the torques since that method is easiest to apply here since the distances and angles are easy to find. We will locate the pivot at the floor because we have two unknowns there.

    r (m) F (N) θ direction τz = rFsinθ
    0 fs MAX - - 0
    0 Nf - - 0
    ½L mg π/2-θ CW -½Lmgsin(π/2-θ)
    L Nw θ CCW LNwsinθ

    Since τz = 0, the equation we get is

    -½Lmgsin(π/2-θ) + LNwsinθ = 0 .

    Eliminating L and noting that sin(π/2-θ) = cosθ yields,

    -½mgcosθ + Nwsinθ = 0 .

    The force equations gave Nw = μsmg, so we have

    -½mgcosθ + μsmgsinθ = 0 .

    Rearranging and using tanθ = sinθ /cosτ, we get

    θ = tan-1(1 / 2μs) .

    If the angle were any smaller than this, the ladder would slip.

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