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Learn/ 9th Physics/ Unit 4 /Numericals

Unit 4: Turning Effects of Force — Numericals

9th Class Physics · Unit 4: Turning Effects of Force

4.1.A force of 200 N is acting on a cart at an angle of 30° with the horizontal direction. Find the x and y-components of the force.
Given
Force F = 200 N
Angle θ = 30°
Formula
X-Component Fx = F cos θ
Y-Component Fy = F sin θ
Working
Fx = (200)(0.8660) = 173.2 N
Fy = (200)(0.5) = 100 N
Result
Components Fx = 173.2 N, quad Fy = 100 N
4.2.A force of 300 N is applied perpendicularly at the knob of a door to open it as shown in the given figure. If the knob is 1.2 m away from the hinge, what is the torque applied? Is it positive or negative torque?
Given
Force F = 300 N
Moment Arm L = 1.2 m
Formula
Torque tau = F × L
Working
tau = (300)(1.2) = 360 N m
Result
Magnitude and Sign tau = 360 N m (positive, anticlockwise)
4.3.Two weights are hanging from a metre rule at the positions as shown in the given figure. If the rule is balanced at its centre of gravity (C.G), find the unknown weight w. The known weight is 4 N at 30 cm from the pivot, and the unknown weight is at 40 cm from the pivot.
Given
Couple arm 1 CA = 30 cm = 0.3 m
Couple arm 2 OB = 40 cm = 0.4 m
Known weight w1 = 4 N
Formula
Principle of Moments w1 × CA = w2 × OB
Working
4 × 0.3 = w2 × 0.4
1.2 = w2 × 0.4
w2 = 1.20.4 = 3 N
Result
Unknown weight w2 = 3 N
4.4.A see-saw is balanced with two children sitting near either end. Child A weighs 30 kg and sits 2 metres away from the pivot, while child B weighs 40 kg and sits 1.5 metres from the pivot. Calculate the total moment on each side and determine if the see-saw is in equilibrium.
Given
Mass of child A m1 = 30 kg
Mass of child B m2 = 40 kg
Distance of child A from pivot d1 = 2 m
Distance of child B from pivot d2 = 1.5 m
Formula
Weight of child A (g = 10 m/s²) w1 = m1 g = 30 × 10 = 300 N
Weight of child B (g = 10 m/s²) w2 = m2 g = 40 × 10 = 400 N
Working
Moment on side A I1 = w1 × d1 = 300 × 2 = 600 N m
Moment on side B I2 = w2 × d2 = 400 × 1.5 = 600 N m
Result
Conclusion I1 = I2 = 600 N m , so the see-saw is in equilibrium
4.5.A crow bar is used to lift a box as shown in the given figure. If the downward force of 250 N is applied at the end of the bar, how much weight does the other end bear? The crowbar itself has negligible weight. The downward force is applied at 30 cm from the pivot, and the box is at 5 cm from the pivot.
Given
Downward force F1 = 250 N
Distance to downward force (OA) OA = 30 cm = 0.3 m
Distance to box (OB) OB = 5 cm = 0.05 m
Formula
Principle of Moments F1 × OA = F2 × OB
Working
F2 = frac{F1 × OA{OB = 250 × 0.30.05
F2 = 750.05 = 1500 N
Result
Weight of block F2 = 1500 N
4.6.A 30 cm long spanner is used to open the nut of a car. If the torque required for it is 150 N m, how much force F should be applied on the spanner?
Given
Moment arm (length of spanner) L = 30 cm = 0.3 m
Required torque tau = 150 N m
Formula
Torque formula tau = F × L
Working
F = tauL = 1500.3
F = 500 N
Result
Force required F = 500 N
4.7.A 5 N ball hanging from a rope is pulled to the right by a horizontal force E. The rope makes an angle of 60° with the ceiling, as shown in the given figure. Determine the magnitude of force F and tension T in the string.
Given
Weight of ball w = 5 N
Angle with ceiling θ = 60°
Formula
Vertical component equilibrium T sin θ = w
Solve for tension T = wsin θ = wsin 60°
Working
T = frac{5}{frac{sqrt{3}{2} = frac{5 × 2}{sqrt{3} = frac{10}{sqrt{3}
T = 5.8 N
Formula
Horizontal component equilibrium T cos θ = F
Working
F = 5.8 × cos 60° = 5.8 × 0.5 = 2.9 N
Result
Forces F = 2.9 N, quad T = 5.8 N
4.8.A signboard is suspended by means of two steel wires as shown in figure. If the weight of the board is 200 N, what is the tension in the strings? The two wires are equidistant from the center of the board, 2 m on each side.
Given
Weight of signboard w = 200 N
Symmetry condition Wires equidistant from center
Formula
Vertical force equilibrium T1 + T2 = w
By symmetry T1 = T2
Working
2T1 = 200
T1 = 2002 = 100 N
Result
Tension in each wire T1 = T2 = 100 N
4.9.One girl of 30 kg mass sits 1.6 m from the axis of a see-saw. Another girl of mass 40 kg wants to sit on the other side, so that the see-saw may remain in equilibrium. How far away from the axis, the other girl may sit?
Given
Mass of girl 1 m1 = 30 kg
Distance of girl 1 from axis d1 = 1.6 m
Mass of girl 2 m2 = 40 kg
Formula
Weight of girl 1 (g = 10 m/s²) w1 = m1 g = 30 × 10 = 300 N
Weight of girl 2 (g = 10 m/s²) w2 = m2 g = 40 × 10 = 400 N
Principle of moments w1 × d1 = w2 × d2
Working
300 × 1.6 = 400 × d2
480 = 400 × d2
d2 = 480400 = 1.2 m
Result
Distance from axis d2 = 1.2 m
4.10.Find the tension in each string of them as shown in given figure, if the block weighs 150 N. The inclined string makes an angle of 60° with the horizontal.
Given
Weight of block w = 150 N
Angle of inclined string θ = 60°
Formula
Vertical component equilibrium T1 sin θ = w
Working
T1 = wsin θ = 150sin 60°
T1 = frac{150}{frac{sqrt{3}{2} = frac{150 × 2}{sqrt{3} = frac{300}{sqrt{3}
T1 ≈ 173.2 N
Formula
Horizontal component equilibrium T2 = T1 cos θ
Working
T2 = 173.2 × cos 60° = 173.2 × 0.5 = 86.6 N
Result
Tensions T1 = 173.2 N, quad T2 = 86.6 N