A heavy crate of mass 46.0 kg is pulled at constant speed by a dockworker who pulls with a 330-N force at an angle ? with the horizontal (see figure below). The magnitude of the friction force between the crate and the pavement is 221 N. What is the angle ? of the rope with the horizontal? What is the magnitude of the normal force exerted by the pavement on the crate?

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A: Thank you for the question. Since, we are entitled to answer up to 3 sub-parts, we’ll answer the…

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A: We can solve the problem by considering the equilibrium condition as below.

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A: Given:-  A woman who weighs  Ww=  602 N She sits on a chair with her feet on the floor and her arms…

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A: Write the given values with suitable variables. m=52 kgFr=354 P=221 Nv=25 cm/s=0.25 m/s Here, m…

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A: Draw the free-body diagram of the block as shown below:

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A: Firstly make the free body diagram and balance forces in horizontal and vertical directions.

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Q: Problem 3: You need to push a box across carpet. The box and its contents have a mass of 75 kg. The…

A: We will first identify direction of forces on the box. Then we calculate magnitude of frictional…

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A: Given, The following forces are acting on a 250 g box on a horizontal surface:  A downward force of…

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A: Given: The mass of the crate is m = 20 kg. The applied force is F = 100 N. The angle is ϴ = 200.

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A: Consider the free-body diagram for the forces on the teddy bear as follows.

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A: Please see the answer below

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A: mass of girl = 50 kg mass of boy = 100 kg distance between girl and boy = 6 m force exerted by boy…

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Q: T1 T2

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A: a. The magnitude of the normal force is N+Fsinθ=mgN=mg-Fsinθ=2.80 kg9.8 m/s2-12.0 Nsin30.0°=21.4 N

Q: a. Find the horizontal and vertical components of the force. b. Is the horizontal component of the…

A: Magnitude of force acting on the strap (F)=40lb a)angle made by force with horizontal =60o cosθ=Fx/F…

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Q: A woman pushes a block of mass m = 25.0 kg up a rough plane sloped at an angle of θ = 20◦ by using…

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Q: Wheaties box from the Cheerios box?

A: Here We apply newton first law   F=ma

Q: net force on the stone

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A: